Question

Assertion :The length of subnormal to $S=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1$ at the point $(x_1,y_1)$ is $\frac{b^2{x}_1}{a^2}$ Reason: The length of tangent to $S=\frac{x^2}{a^2}+\frac{y^2}{b^2}-1$ from the point $(x_1,y_1)$ is $=x_1-\frac{a^2}{x_1}.$

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Both Assertion and Reason are incorrect

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

For a standard Hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

The equation of normal at some points let's say $P(x_1,y_1)$ is given as:

$\Rightarrow a^2{y}_1(x-x_1)+b^2{x}_1(y-y_1)=0$ ...$\left(1\right)$

As you can see in the figure the normal at the point $P$ meets the $x-$ axis at the point $N$.

At point $N,$ the $y$ - coordinate is Zero. Putting value of $(y=0)$ in equation $\left(1\right)$

$\Rightarrow a^2{y}_1(x-x_1)+b^2{x}_1(0-y_1)=0$

$\Rightarrow x=\frac{b^2{x}_1}{a^2}+x_1$

Hence point $N$ $\Rightarrow (\frac{b^2{x}_1}{a^2}+x_1,0)$

$PG$ is the perpendicular drwn from point $P$ to the $x-$ axis, it meets $x-$axis at $G$, the coordinates of $G$ are $(x_1,0)$

The length of sub-normal of a hyperbola is actually the projection of normal on the $x-$ axis. From the figure you can see the subnormal is $GN$.

$\Rightarrow GN=\frac{b^2{x}_1}{a^2}+x_1-x_1=\frac{b^2{x}_1}{a^2}$

Hence $,GN=\frac{b^2{x}_1}{a^2}$

The equation of tangent $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ at some points let's say $P(x_1,y_1)$ is given as:

$\Rightarrow \frac{x{x}_1}{a^2}-\frac{y{y}_1}{b^2}=1$ ...$\left(1\right)$

As you can see in the figure the Tangent at point $P$ meets the $x-$ axis at point $T$.

At point $T,$ the $y$ - coordinate is Zero. Putting value of $y=0$ in equation $\left(1\right)$

$\Rightarrow \frac{x{x}_1}{a^2}-\frac{y\left(0\right)}{b^2}=1$

$\Rightarrow x=\frac{a^2}{x_1}$

Hence point $T$ is $(\frac{a^2}{x_1},0)$

$PG$ is the perpendicular drwn from point $P$ to the $x-$ axis, it meets $x-$axis at $G$, the coordinates of $G$ are $(x_1,0)$

Hence $TG=x_1-\frac{a^2}{x_1},$ and $PG=y_1$

$\Rightarrow$ The length of the tangent is $PT=\sqrt{(PG{)}^2+(TG{)}^2}$

$PT=\sqrt{y_1^2+{(x_1-\frac{a^2}{x_1})}^2}$

As the assertion is correct but reason is incorrect, and correct answer is $C$.