Statement-1: Consider the parabola (y-2)2=8(x-1) and circle (x+5)2+(y-2)2=8. There is exactly one point such that pair of tangents to both parabola and circle drawn from it are perpendicular. Statement-2: Director circles of parabola and the given circle can intersect in atmost two points.

Question

Statement-1: Consider the parabola (y-2)2=8(x-1) and circle (x+5)2+(y-2)2=8. There is exactly one point such that pair of tangents to both parabola and circle drawn from it are perpendicular. Statement-2: Director circles of parabola and the given circle can intersect in atmost two points. (A) Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1. (B) Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1. (C) Statement-1 is true, statement-2 is false. (D) Statement-1 is false, statement-2 is true.

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/ee3092b575d23820c34e7f2955767e7d-.png" /> Equation of the director circle of the circle (x+5)2+(y-2)2 (x+5)2+(y-2)2=16 and of (y-2)2=8(x-1) is x=-1. Clearly the line x=-1 touches (x+5)2+(y-2)2=16 Hence only one such point exist.

Q. Statement-1: Consider the parabola (y−2)2=8(x−1) and circle (x+5)2+(y−2)2=8. There is exactly one point such that pair of tangents to both parabola and circle drawn from it are perpendicular. Statement-2: Director circles of parabola and the given circle can intersect in atmost two points.

Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.

Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.

Statement-1 is true, statement-2 is false.

Statement-1 is false, statement-2 is true.

Equation of the director circle of the circle (x+5)2+(y−2)2 (x+5)2+(y−2)2=16 and of (y−2)2=8(x−1) is x=−1. Clearly the line x=−1 touches (x+5)2+(y−2)2=16 Hence only one such point exist.

Equation of the director circle of the circle $(x + 5)^{2} + (y - 2)^{2}$
$(x + 5)^{2} + (y - 2)^{2} = 16$ and of $(y - 2)^{2} = 8 (x - 1)$ is $x = - 1$ .
Clearly the line $x = - 1$ touches $(x + 5)^{2} + (y - 2)^{2} = 16$
Hence only one such point exist.