1. Tardigrade
  2. Question
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  4. 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.

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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.

Conic Sections

Solution:

image
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.