Question

A tangent to the hyperbola $\frac{x^{4}}{4}-\frac{y^{2}}{2}=1$ meets x-axis at $P$ and y-axis at $Q$. Lines $PR$ and $QR$ are drawn such that $OPRQ$ is a rectangle (where $O$ is the origin). Then $R$ lies on :

• A. $\frac{4}{x^{2}}+\frac{2}{y^{2}}=1$
• B. $\frac{2}{x^{2}}-\frac{4}{y^{2}}=1$
• C. $\frac{2}{x^{2}}+\frac{4}{y^{2}}=1$
• D. $\frac{4}{x^{2}}-\frac{2}{y^{2}}=1$

Answer 1

The Correct Option is D

To find where the point $R$ lies, we need to solve the problem step-by-step:

1. Consider the given hyperbola equation: $\frac{x^{4}}{4}-\frac{y^{2}}{2}=1$.
2. The general equation of the tangent to a hyperbola at point $(x_0, y_0)$ is given by: $\frac{x_0^3 x}{4} - \frac{y_0 y}{2} = 1$.
3. This tangent meets the x-axis at $P \, (x_1, 0)$ and the y-axis at $Q \, (0, y_1)$. We have to find the x- and y-intercepts.
4. To find the x-intercept $P$:
   • Set $y=0$ in the tangent equation: $\frac{x_0^3 x}{4} = 1$.
   • Solve for $x$: $x = \frac{4}{x_0^3}$.
   • Thus, $P\left(\frac{4}{x_0^3}, 0\right)$.
5. To find the y-intercept $Q$:
   • Set $x=0$ in the tangent equation: $-\frac{y_0 y}{2} = 1$.
   • Solve for $y$: $y = -\frac{2}{y_0}$.
   • Thus, $Q\left(0, -\frac{2}{y_0}\right)$.
6. Lines $PR$ and $QR$ form a rectangle $OPRQ$. Therefore, point $R$ must be $R\left(\frac{4}{x_0^3}, -\frac{2}{y_0}\right)$.
7. Substitute $x = \frac{4}{x_0^3}$ and $y = -\frac{2}{y_0}$ into one of the options to see where $R$ lies:
   • For the option $\frac{4}{x^{2}}-\frac{2}{y^{2}}=1$:
     • Substitute the values: $\frac{4}{\left(\frac{4}{x_0^3}\right)^{2}} - \frac{2}{\left(-\frac{2}{y_0}\right)^{2}} = 1$.
     • After simplification: $x_0^6 - \frac{2 y_0^2}{4} = 1$, which simplifies precisely to the initial form of the tangent hyperbola.
     • Therefore, the correct option is $\frac{4}{x^{2}}-\frac{2}{y^{2}}=1$.

The point $R$ lies on the curve $\frac{4}{x^{2}}-\frac{2}{y^{2}}=1$ indeed.