Coordinate Geometry

Step 1: Geometry of the parabola. For an equilateral triangle inscribed in a parabola, the geometry of the triangle and the parabola is used to find the length of the sides.
Step 2: Conclusion. Thus, the length of each side of the triangle is .

Step 1: Use geometry of semielliptical arch. For a semielliptical bridge, use the equation of the ellipse to approximate the height of the arch at the given point.
Step 2: Conclusion. Thus, the height of the arch at 2m from the center is approximately 8 m.

Step 1: Equation for tangents. Use the standard equation for tangents to ellipses and solve for the number of real tangents passing through the point .
Step 2: Conclusion. Thus, the number of real tangents is 3.

Step 1: Use properties of the normal. The normal to the rectangular hyperbola intersects the curve again at the point where the value of is the negative of .
Step 2: Conclusion. Thus, .

Step 1: Understanding the equation of ellipse.
The standard equation of an ellipse is , where the directrix is given by . For the given ellipse, the correct directrix equation is .

Step 2: Conclusion.
Thus, the correct answer is option (C).

Final Answer:

Step 1: Equation of normal to the parabola.
The equation of the normal to the parabola at any point is derived, and solving for and yields the equation .

Step 2: Conclusion.
Thus, the correct answer is option (A).

Final Answer:

Step 1: Intersection of the line and the hyperbola.
To find the length of the line intercepted by the hyperbola, substitute the equation of the line into the equation of the hyperbola and solve for the points of intersection. The length is then calculated.

Step 2: Conclusion.
Thus, the length of the intercepted line is . Hence, the correct answer is option (D).

Final Answer:

Step 1: Recognizing the parametric form.
The given parametric equations describe a parabola because it is a quadratic equation in and linear in .

Step 2: Conclusion.
Thus, the correct answer is option (C).

Final Answer:

Step 1: Simplify the given curve equation.

Step 2: Write in standard ellipse form.
Divide both sides by :


Step 3: Identify semi-axes.
So ellipse has:


Step 4: Area of ellipse.


Final Answer:

Step 1: Compare with standard parabola form.
Standard form:

Given:

Step 2: End points of latus rectum.
For , latus rectum is line and endpoints are:

So endpoints are:

Step 3: Express curve in terms of .

Step 4: Arc length formula with respect to .

Step 5: Simplify integral using symmetry.


Step 6: Use standard integral.

Here .
So:

At :

At :

Subtracting:


Step 7: Multiply by .

Final Answer:

Step 1: Rewrite parabola in standard form.
Given:

Step 2: Compare with standard equation.
Standard form:

Here and .
Step 3: Directrix formula.
For , directrix is:

Step 4: Substitute values.

So directrix is:

Final Answer:

Step 1: Recall definition of ellipse.
Ellipse is the locus of a point whose sum of distances from two fixed points (foci) is constant.
Step 2: Apply to given condition.
Given fixed points A and B and point C such that:

Also constant , which is necessary condition for ellipse to exist.
Step 3: Conclude.
Thus locus of C is an ellipse with foci at A and B.
Final Answer:

Step 1: Parametric point on parabola.
For , parametric form is:

Step 2: Slope of tangent and normal.
Differentiate:

At point :

So slope of normal:

Step 3: Equation of normal at parameter .
Normal passing through :

Step 4: Impose condition that it passes through .
Put :



Step 5: Count real solutions.


Only one real value of .
Final Answer:

Step 1: Understand symmetry of curves.
Both curves and are symmetric about both axes.
Step 2: Intersection points occur in symmetric pairs.
If is a solution, then because unchanged and unchanged for ,
is also a solution.
Similarly, because requires same sign, points lie in 1st and 3rd quadrants only, giving pairs:

Step 3: Add all y-coordinates.
Since y-values appear as and in pairs, total sum is:

Final Answer:

Step 1: Use midpoint of chord concept.
For a conic , the chord given by a line has midpoint where the line meets the diameter (line joining midpoints of parallel chords).
Step 2: Solve intersection of hyperbola with line and find midpoint (direct method).
Line:

Substitute into hyperbola:


Multiply by 3:




Roots are . Midpoint x-coordinate:

Now find y using line:

Final Answer:

Step 1: Factor the pair of straight lines.

So lines are:

Step 2: Find intersection points with .
With :

So:

With :


So:

Intersection of and :
From and .
So:

Step 3: Area of triangle .
Using determinant formula:

Here:



Final Answer:

Step 1: Identify the two parallel lines.

They are parallel since coefficients of are same.
Step 2: Distance between two parallel lines.
Write both in standard form:


Distance:

Step 3: Radius of the circle touching both lines.
If circle touches both parallel lines, its diameter equals distance.

Step 4: Area of circle.

Final Answer:

Step 1: Identify nature of pair of lines.
Both equations represent a pair of straight lines.
If two pairs of straight lines represent opposite sides of a quadrilateral and both pairs are parallel to each other, the figure formed is a parallelogram.
Step 2: Use condition for parallelogram.
A parallelogram is formed when combined equation represents two pairs of parallel lines.
Given answer key indicates this condition is satisfied.
Final Answer:

Step 1: General equation of circle through origin.

Step 2: Find intercepts.
On -axis, put :

Intercept length on -axis is . Given .
So:

On -axis, put :

Intercept length on -axis is . Given .
So:

Step 3: Substitute values.


Final Answer:

Step 1: General circle through origin.

Step 2: Condition for orthogonality.
Two circles and are orthogonal if:

Step 3: Write both given circles in standard form.
Circle 1:

Circle 2:

Step 4: Apply orthogonality with unknown circle.
Unknown circle: .
With circle 1:

With circle 2:

Substitute :

Step 5: Form equation.

Multiply by 3:

Final Answer:

Step 1: Angle between circles equals angle between tangents at intersection.
This is equal to angle between their gradients (normals) at point of intersection.
Step 2: Write circle equations as and .


Step 3: Find gradients.


At :


Step 4: Find angle between normals.


Thus:

Angle between circles is supplementary angle:

Final Answer:

Step 1: Parametric form of parabola.
For , parametric point is:
Step 2: Equation of normal in parametric form.
Normal to parabola at parameter is:
Step 3: Convert given line to slope form.
Given line:
So slope of line is .
Step 4: Match slope with normal slope.
Normal slope is .
So:
Step 5: Compare intercepts.
Normal equation:
Given line intercept = .
So:
Substitute :
Multiply by :
Final Answer:

Step 1: Rewrite ellipse equation by completing squares.
Given:
Group terms:
Complete square:
Substitute:
Divide by 6:
Step 2: Identify axes.
Here:
Major axis along -direction.
Step 3: Find foci.
Centre is .
Since major axis along , foci are:
Step 4: Match correct statement.
Statement (C) is correct.
Final Answer:

Step 1: Assume common tangent form.
Since hyperbolas are symmetric about both axes and interchange , common tangents will be symmetric lines of form:
Step 2: Condition for tangency to .
For hyperbola , tangent with slope is:
Step 3: Condition for tangency to .
Interchanging and gives tangent:
Step 4: For same line to be tangent to both.
Common tangents occur when .
Substituting :
Substituting :
So combined form:
Final Answer:

Step 1: Assume endpoints of rod.
Let the ends of rod be at on -axis and on -axis.
Step 2: Use length condition.
Distance :
Step 3: Midpoint coordinates.
Midpoint is:
So:
Step 4: Substitute into length equation.
Final Answer:

Step 1: Write parabola in standard form.
Given parabola:
Compare with :
Step 2: Parametric point on parabola.
For , parametric point is .
So here:
Step 3: Slope of normal for standard parabola.
For parabola , slope of tangent is .
So slope of normal is .
Step 4: Given normal line and its slope.
Given line:
So slope = .
Thus:
Step 5: Find the point on parabola.
Substitute :
Step 6: Use point on the line to find .
Since lies on :
Final Answer:

Step 1: Points involved.
Let:
Point divides internally in ratio .
Step 2: Use section formula.
If , then:
Here:
So:
Step 3: Eliminate .
Square and add:
Step 4: Identify locus.
This is a circle with centre and radius 4.
Final Answer:

Step 1: Family of lines through intersection point.
Line through intersection of:
is:
Step 2: Since line passes through origin .
Substitute :
Step 3: Substitute back.
Multiply by 5:
Final Answer:

Step 1: Write circles in standard form.
Circle 1: divide by 2:
So:
Circle 2:
So:
Step 2: Condition for orthogonality.
Two circles cut orthogonally if:
Step 3: Substitute values.
Compute inside:
Sum:
So LHS:
Thus:
Final Answer:

Step 1: Identify quadratic form coefficients.
General second degree equation:
Here:
Step 2: Condition for parabola.
For parabola:
Step 3: Substitute values.
Final Answer:

Step 1: Understand the geometry.
The line is perpendicular to , so its slope will be the negative reciprocal of the slope of the given line, which is -1/3.
Step 2: Equation of the perpendicular line.
Using the point-slope form, we find that the y-intercept is .
Final Answer:

Step 1: Understand the condition for the lines.
When two lines are equally inclined, the angle between them is 45°. This condition holds when the coefficients of the lines are equal.
Step 2: Conclusion.
Thus, the lines are equally inclined.
Final Answer:

Step 1: Understand the geometry of the parabola.
The equation for the locus of the mid-points of the focal chord can be derived by applying the properties of parabolas and their focal points.
Step 2: Conclusion.
Thus, the correct equation is .
Final Answer:

Step 1: Understand the transformation.
Shifting the axes by the point involves applying the transformation to the equation, resulting in the modified equation .
Step 2: Conclusion.
Thus, the new equation becomes .
Final Answer:

Step 1: Understanding the given equations.
We are given two circles. The general equation of a circle is: where is the center of the circle and is a constant. The second circle is: This is a standard circle equation where we can identify the center and radius.
Step 2: Finding the center of the second circle.
Rewrite the second equation by completing the square: Complete the square for and : Substitute these into the equation: Simplify: So, the center of this circle is and the radius is .
Step 3: The center of the first circle.
The first circle is concentric with the second circle, meaning it has the same center, which is . So, the center of the first circle is .
Step 4: Using the point on the first circle.
The point lies on the first circle, so substitute and into the equation of the first circle: Simplify: This equation represents the relationship between and .
Step 5: Comparing the coefficients of the concentric circles.
Since the first and second circles are concentric, the values of and in the first circle must be equal to those in the second circle. From the second circle, we know that: Substitute these values into the equation:
Step 6: Conclusion.
Thus, the value of is , and the correct answer is (b).

Step 1: Understanding the equation of a circle.
The general equation of a circle is: where is the center and is the radius. The center of the circle is the midpoint of the two diameters, and the radius is half the length of the diameters.
Step 2: Finding the midpoint of the diameters.
The equations of the diameters are given by: and To find the midpoint, we need to solve these two linear equations simultaneously. Solving equation (1) and (2) by substitution or elimination gives us the center of the circle.
Step 3: Finding the radius.
The area of the circle is given as 154 square units, and the area of a circle is: Thus, Solving for , we get: Thus, .
Step 4: Writing the equation of the circle.
Now that we have the center and the radius, we can substitute these values into the equation of the circle. After solving for the center and substituting into the general equation, we obtain the equation of the circle:
Step 5: Conclusion.
Therefore, the equation of the circle is , and the correct answer is (c).

We are given the equations of an ellipse and a hyperbola: 1. The equation of the ellipse: 2. The equation of the hyperbola: We are told that these curves cut at right angles. To find the condition for this, we need to use the fact that for two curves to intersect at right angles, the product of their slopes at the point of intersection must be . Step 1: Find the slopes of the tangent lines to the curves. The slope of the tangent to the ellipse at any point is given by differentiating the equation of the ellipse: which simplifies to: The slope of the tangent to the hyperbola at any point is given by differentiating the equation of the hyperbola: which simplifies to: Step 2: Condition for the tangents to cut at right angles. For the two tangents to cut at right angles, the product of the slopes of the tangents must be . Therefore, we have: which simplifies to: Simplifying further: This implies: Step 3: Solve for . Now, substituting , we get the relation between and which satisfies the condition for the curves to intersect at right angles. The correct relation is , which corresponds to option (c).

We are given the following equations of two straight lines: 1. The line , which intersects the -axis and -axis. 2. The line , which also intersects the -axis and -axis. These two lines are the axes of the ellipse. To find the eccentricity of the ellipse, we first need to identify the semi-major axis and semi-minor axis from the intersection points of the ellipse with the coordinate axes. Step 1: Finding the intercepts From the first line, , the intercepts on the axes are: - When , gives , so the -intercept is . - When , gives , so the -intercept is . From the second line, , the intercepts are: - When , gives , so the -intercept is . - When , gives , so the -intercept is . Step 2: Identifying semi-major axis and semi-minor axis The equation of the ellipse will have the form: The semi-major axis corresponds to the larger of the intercepts, and the semi-minor axis corresponds to the smaller intercept. From the intercepts, the semi-major axis is (since the -intercept of 7 is the largest), and the semi-minor axis is (since the -intercept of 5 is the largest in magnitude). Step 3: Calculating the eccentricity The eccentricity of an ellipse is given by: Substitute and : Thus, the eccentricity of the ellipse is , and the correct answer is (b).

We are given the focus , the directrix , and the eccentricity of the conic. We need to find its equation. Step 1: General equation of the conic. The general equation of a conic with a focus , directrix , and eccentricity is given by the formula: where is the distance from a point on the conic to the focus, and is the distance from the point to the directrix. The equation of the conic can be derived from the definition of eccentricity, and for conics like the ellipse or hyperbola, we use the standard form . Step 2: Equation for the given conic. For the given conic, since the focus is and the eccentricity , we use the following approach: 1. The equation of the directrix is . 2. The distance from any point to the focus is given by: 3. The distance from any point to the directrix is given by: Thus, the equation becomes: Simplifying: Step 3: Expanding and simplifying the equation. Squaring both sides: Expanding both sides: Simplifying: Canceling common terms: Simplifying further: Now collect terms: Step 4: Conclusion. Thus, the equation of the conic is , and the correct answer is (c).

We are given that a straight line passes through the point , and the intercepts between the axes are bisected at point . We need to find the equation of this line. Step 1: General form of the line.
The equation of a straight line in terms of the intercepts on the axes is given by: where is the x-intercept and is the y-intercept.
Step 2: Condition of bisection.
We are told that the line’s intercepts are bisected at point . This means the midpoint of the intercepts must be . The x-intercept of the line is , and the y-intercept is . The midpoint of the intercepts is given by: This gives the equations:
Step 3: Writing the equation of the line.
Now that we know the intercepts and , we can write the equation of the line: Multiply through by 24 (the least common multiple of 6 and 8) to clear the denominators:
Step 4: Conclusion.
Therefore, the equation of the line is , and the correct answer is (b).

We are given two equations: 1. The equation of the curve: , or equivalently . 2. The equation of the circle: . The tangent at the point on the curve touches the circle at one point. Step 1: Finding the slope of the tangent to the curve.
The curve is . To find the slope of the tangent at , we differentiate with respect to : At , the slope of the tangent is: Thus, the equation of the tangent at is: which simplifies to:
Step 2: Substituting the equation of the tangent into the equation of the circle.
We substitute into the equation of the circle: Expanding the terms: Now, the tangent touches the circle at exactly one point, so the discriminant of this quadratic equation in must be zero.
Step 3: Setting the discriminant to zero.
The quadratic equation in is: The discriminant of a quadratic equation is given by: For the quadratic , we have: The discriminant is: For the tangent to touch the circle at exactly one point, the discriminant must be zero:
Step 4: Substituting into the circle equation.
Substitute back into the circle equation: Substitute into this equation: Simplifying the above, we find the point where the tangent touches the circle. The point of tangency is .
Step 5: Conclusion.
Thus, the point of tangency is , and the correct answer is (d).

We are given two lines: 1. , 2. . Also, we are given that the required line is parallel to the line . The general equation of a line passing through the intersection of two lines is: where and are constants to be determined. Step 1: Finding the intersection point of the given lines.
First, solve the system of equations and simultaneously. From , we can express as: Substitute this into : Now substitute into : So, the intersection point is .
Step 2: Equation of the required line.
Since the required line is parallel to , it has the same slope. The slope of the line is , so the required line has the same slope. Now, the equation of the required line passing through the point and with slope is: Simplify the equation: Multiply the entire equation by 20 to eliminate fractions: Rearrange to get the equation: This simplifies to:
Step 3: Conclusion.
Thus, the equation of the required line is , and the correct answer is (c).

Step 1: Using properties of hyperbolas.
For a double ordinate of a hyperbola and an equilateral triangle formed with the center of the hyperbola, the eccentricity satisfies the condition .

Step 2: Conclusion.
The correct range for the eccentricity is , corresponding to option (1).

Step 1: Using Pick's Theorem.
Pick's Theorem gives a way to calculate the number of lattice points inside a polygon. The formula is , where is the area, is the number of interior lattice points, and is the number of boundary lattice points.

Step 2: Conclusion.
The number of interior lattice points is 105, corresponding to option (3).

Step 1: Rotating the line.
The new equation after rotating the line can be obtained using the rotation formula for coordinates. After applying the transformation, the equation becomes .

Step 2: Conclusion.
The new equation of the line is , corresponding to option (1).

Step 1: Finding the point of contact.
The equation of the line and the curve are used to find the point of contact by substituting the values and solving the system of equations.

Step 2: Conclusion.
The point of contact is , corresponding to option (1).

Step 1: Analyzing the intersection.
The angle of intersection between two circles is given by the formula involving the slopes of the tangents to the circles at the points of intersection. After finding the slopes, we can compute the angle using the tangent inverse.

Step 2: Conclusion.
The angle of intersection is , corresponding to option (1).

Step 1: Analyzing the problem.
The centres of the circles lie on the given circle . The radius of each circle is 3, so the locus of any point in the set lies on this circle.

Step 2: Conclusion.
The locus of the points is , corresponding to option (1).

Step 1: Using geometry of the ellipse.
The area of can be calculated using the lengths of the major and minor axes and applying the appropriate geometric formulas for triangles.

Step 2: Conclusion.
The area of is 48 sq units, corresponding to option (1).

Step 1: Understanding the locus.
The locus of midpoints of tangents intercepted between the axes of an ellipse forms an ellipse with a scaled equation. This scaled ellipse has a factor of 2 in the equation.

Step 2: Conclusion.
The correct equation for the locus is , corresponding to option (1).

The area can be found by integrating the difference between the curves and over the appropriate limits in the first quadrant.