# NCERT Class 11-Math's: Exemplar Chapter –11 Conic Sections Part 5

Glide to success with Doorsteptutor material for ISAT : Get full length tests using official NTA interface: all topics with exact weightage, real exam experience, detailed analytics, comparison and rankings, & questions with full solutions.

**Question 13:**

The area of the triangle formed by the lines joining the vertex of the parabola to the ends of its latus rectum is

(A) . units

(B) . units

(C) . units

(D) . units

**Answer:**

The correct option is (C).

From the figure, OPQ represent the triangle whose area is to be determined. The area of the triangle

**Question 14:**

The equations of the lines joining the vertex of the parabola to the points on it which have abscissa are

(A)

(B)

(C)

(D)

**Answer:**

(B) is the correct choice.

Let P and Q be points on the parabola and OP, OQ be the lines joining the vertex O to the points P and Q whose abscissa are 24.

Thus

or .

Therefore the coordinates of the points P and Q are and respectively. Hence the lines are

**Question 15:**

The equation of the ellipse whose centre is at the origin and the *x*-axis, the major axis, which passes through the points and is

(A)

(B)

(C)

(D)

**Answer:**

(B) is the correct choice.

Let be the equation of the ellipse. Then according to the given conditions, we have

Which gives

Hence, required equation of ellipse is .

**Question 16:**

The length of the transverse axis along *x*-axis with centre at origin of a hyperbola is and it passes through the point . The equation of the hyperbola is

(A)

(B)

(C)

(D) None of these

**Answer:**

(C) Is the correct choice.

Let represent the hyperbola. Then according to the given condition, the length of transverse axis, i.e., .

Also, the point lies on the hyperbola, so, we have

which gives

Hence, the equation of the hyperbola is