Question

A standard equation of conic satisfies the point $(2,4)$ and the conic is such that the segment of any of its tangents at any point contained between the coordinate axes is bisected at the point of tangency. Then

• A. equation of directrices of conic are $x+y=\pm 4$
• B. eccentricity of conic $=\sqrt{2}$
• C. the foci of the conic are $(4,4)\text{and}(-4,-4)$
• D. tangent equation at $(2,4)$ to conic is $4x+2y=16$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A equation of directrices of conic are $x+y=\pm 4$
B eccentricity of conic $=\sqrt{2}$
C the foci of the conic are $(4,4)\text{and}(-4,-4)$
D tangent equation at $(2,4)$ to conic is $4x+2y=16$

Let at point $(x_1,y_1)$ on the conic tangent is drawn and let tangent meet the axes at $(a,0)$ and $(0,b)$
Now according to the question
$a=2x_1,b=2y_1$
Hence equation of tangent will be
$\frac{x}{2x_1}+\frac{y}{2y_1}=1\Rightarrow x{y}_1+y{x}_1=2x_1{y}_1$
which is of the form tangent at $(x_1,y_1)$ on the conic $xy=c^2$
i.e., $x{y}_1+y{x}_1=2c^2$
or $x{y}_1+y{x}_1=2x_1{y}_1[\because (x_1,y_1\left)\text{lies on hyperbola}\right]$
Hence tangent at $(2,4)$ is $4x+2y=16$
As the curve passes through $(2,4)$
$\therefore xy=8$
Hence the eccentricity $=\sqrt{2}$

and coordinates of foci are $(\pm \sqrt{2}c,\pm \sqrt{2}c)\equiv (4,4),(-4,-4)$

equation of directrix is $x+y=\pm \sqrt{2}c$
Hence, its equations $x+y=\pm 4$