Question

Two tangents on a parabola are $x-y=0$ and $x+y=0.$ Let $(2,3)$ is the focus of the parabola, then

• A. The equation of the tangent at vertex is $4x-6y+5=0$
• B. The equation of the tangent at vertex is $4x-6y+1=0$
• C. The length of the latus rectum of the parabola is $\frac{10}{\sqrt{52}}$
• D. The length of the latus rectum of the parabola is $\frac{10}{\sqrt{13}}$

Answer 1

Answer: (A), (D)

The correct options are
A The equation of the tangent at vertex is $4x-6y+5=0$
D The length of the latus rectum of the parabola is $\frac{10}{\sqrt{13}}$

The foot of the perpendicular from the focus upon any tangent lies on the tangent at the vertex.


The foot of perpendicular of $(2,3)$ on the line $x-y=0$ is $A(\frac{5}{2},\frac{5}{2})$.
The foot of of perpendicular of $(2,3)$ on the line $x+y=0$ is $B(-\frac{1}{2},\frac{1}{2})$.

The equation of tangent at vertex (Eq. of $AB$) is $4x-6y+5=0$

Let the length of the latus rectum be $4a$
Then the length of perpendicular to the tangent at vertex from the focus is $a.$

$a=\frac{|4\left(2\right)-6\left(3\right)+5|}{\sqrt{52}}\Rightarrow 4a=\frac{10}{\sqrt{13}}$