## JEE Advanced Chapter-wise Archive – Parabola

Q.1₂₀₁₅ Let the curve C be the mirror image of the parabola y² = 4x with respect to the line x + y + 4 = 0. If A and B are the points of intersection of C with the line y = -5 then the distance between A and B is…

Solution:
Mirror image of the parabola y² = 4x with respect to the line x + y + 4 = 0 is C: (x + 4)² = -4(y + 4)
Now the points of intersection of C with the line y = -5 are A(-2, -5) and B(-6, -5)
⇒ AB = |-2 + 6| = |4| = 4

Q.2₂₀₁₅ If the normals of the parabola y² = 4x drawn at the end points of its latus rectum are tangents to the circle (x – 3)² + (y + 2)² = r², then the value of r² is…

Solution:
Equation of the normals are x + y = 3 and x – y = 3
distance from the centre (3, -2) of the circle on both normals is the radius (r) of the circle
⇒ r = 2 ⇒ r² = 4

Q.3₂₀₁₂ Let S be the focus of the parabola y² = 8x and let PQ be the common chord of the circle x² + y² – 2x + 4y = 0 and the given parabola. The area of the triangle PQS is…

Solution:
Let (2t², 4t) be any point on the parabola and S(2, 0) be the focus of the parabola
solving (2t², 4t) with the circle x² + y² – 2x + 4y = 0
⇒ (2t²)² + (4t)² – 2(2t²) + 4(4t) = 0 ⇒ t = 0, 1
⇒ the coordinates of P and Q are (0, 0) and (2, 4)
⇒ The area of the triangle PQS is (2 x 4)/2 = 4

Q.4₁₉₉₄ Point of intersection of tangents at the ends of the latus rectum of the parabola y² = 4x is…

Solution:
Since the point of intersection of tangents at the ends of the latus rectum of the parabola lies on the directrix and remain on the axis of the parabola
⇒ The required point of intersection is (-1, 0)