JEE Mathematics 2023 Question with Solution

Given: The vertices are $({\pm}6,0)$, so $a=6$ and $a^2=36$. The eccentricity is $e=\frac{\sqrt{5}}{2}$.

Find: The value of $d^2$, where $d$ is the length of the segment of the normal between the hyperbola and the $y$-axis.

For the hyperbola

we use the standard parametric point

so here

Using

we get

Hence the hyperbola is

The equation of the normal at parameter $\theta$ is

Its slope is

The given line $\sqrt{2}x+y=2\sqrt{2}$ has slope

Since the normal is parallel to this line,

So,

which gives

for the point in the first quadrant.

Substituting $\theta=\frac{\pi}{4}$ in the normal,

Thus the normal meets the $y$-axis at

Also,

Therefore,

Hence, the required value is $216$.

Given: $a=6$ and $e=\frac{\sqrt{5}}{2}$.

Find: The square of the segment length on the normal from the point of contact to the $y$-axis.

First compute $b$ from eccentricity:

So the hyperbola is

For the point $P=(a\sec\theta,b\tan\theta)$ on the hyperbola, the normal is

With $a=6, b=3$, this becomes

The slope of this line is

Since the normal is parallel to $\sqrt{2}x+y=2\sqrt{2}$, its slope is $-\sqrt{2}$. Hence

In the first quadrant,

Now

The normal becomes

so on the $y$-axis $x=0$ and therefore

Finally,

Therefore, the answer is $216$.