Question

If $4{{x}^{2}}+p{{y}^{2}}=45$ and ${{x}^{2}}-4{{y}^{2}}=5$ cut orthogonally, then the value of $p$
A. $\dfrac{1}{9}$
B. $\dfrac{1}{3}$
C. $3$
D. $18$
E. $9$

Answer 1

Hint: To solve this question we will use the condition where product of the two slopes is $-1$ when both cuts each other orthogonally. We will first calculate the slopes of both the curves by differentiating both the equations. Then we will derive an equation using relation \[{{m}_{1}}{{m}_{2}}=-1\]in terms of ${{x}^{2}}$ and ${{y}^{2}}$. We will solve both the given equations of curves using substitution method and determine the value of ${{x}^{2}}$ and ${{y}^{2}}$. We will then substitute the value of ${{x}^{2}}$ and ${{y}^{2}}$ in the derived equation using relation \[{{m}_{1}}{{m}_{2}}=-1\] and find the value of $p$.

Formula Used: Slope of a curve $m=\dfrac{dy}{dx}$.

Complete step by step solution: We are given equation of two curves $4{{x}^{2}}+p{{y}^{2}}=45$ and ${{x}^{2}}-4{{y}^{2}}=5$ cutting orthogonally and we have to find the value of $p$.
We will first calculate the slope ${{m}_{1}}$of first curve $4{{x}^{2}}+p{{y}^{2}}=45$ by differentiating it with respect to $x$.
\[\begin{align}
  & 4{{x}^{2}}+p{{y}^{2}}=45 \\
 & 4.2x+p.2y\dfrac{dy}{dx}=0 \\
 & 8x+2py\dfrac{dy}{dx}=0 \\
 & \dfrac{dy}{dx}=\dfrac{-8x}{2py} \\
 & \dfrac{dy}{dx}=\dfrac{-4x}{py}
\end{align}\]
The slope of first curve will be ${{m}_{1}}=\dfrac{-4x}{py}$.
We will now calculate the slope ${{m}_{2}}$of first curve ${{x}^{2}}-4{{y}^{2}}=5$ by differentiating it with respect to $x$.
$\begin{align}
  & {{x}^{2}}-4{{y}^{2}}=5 \\
 & 2x-4.2y\dfrac{dy}{dx}=0 \\
 & 2x-8y\dfrac{dy}{dx}=0 \\
 & 8y\dfrac{dy}{dx}=2x \\
 & \dfrac{dy}{dx}=\dfrac{x}{4y}
\end{align}$
The slope of second curve will be ${{m}_{2}}=\dfrac{x}{4y}$.
Now we know that if two curves cut each orthogonally or perpendicularly then product of their slopes is equals to $-1$. So,
\[\begin{align}
  & {{m}_{1}}{{m}_{2}}=-1 \\
 & \dfrac{x}{4y}\times \dfrac{-4x}{py}=-1 \\
 & {{x}^{2}}=p{{y}^{2}}...(i)
\end{align}\]
We will now solve both the given equations of curves $4{{x}^{2}}+p{{y}^{2}}=45$ and ${{x}^{2}}-4{{y}^{2}}=5$.
$\begin{align}
  & {{x}^{2}}-4{{y}^{2}}=5 \\
 & {{x}^{2}}=5+4{{y}^{2}}...(ii)
\end{align}$
Substituting the value of equation(ii) in $4{{x}^{2}}+p{{y}^{2}}=45$.
$\begin{align}
  & 4{{x}^{2}}+p{{y}^{2}}=45 \\
 & 4(5+4{{y}^{2}})+p{{y}^{2}}=45 \\
 & 20+16{{y}^{2}}+p{{y}^{2}}=25 \\
 & {{y}^{2}}(16+p)=25 \\
 & {{y}^{2}}=\dfrac{25}{16+p}.....(iii)
\end{align}$
We will now determine the value of ${{x}^{2}}$ by putting the value of ${{y}^{2}}$ in equation ${{x}^{2}}-4{{y}^{2}}=5$.
\[\begin{align}
  & {{x}^{2}}-4{{y}^{2}}=5 \\
 & {{x}^{2}}-\dfrac{100}{16+p}=5 \\
 & {{x}^{2}}=5+\dfrac{100}{16+p} \\
 & {{x}^{2}}=\dfrac{180+5p}{16+p}...(iv)
\end{align}\]
Substituting the value of equation (iii) and (iv) in equation (i) to determine the value of $p$.
$\begin{align}
  & {{x}^{2}}=p{{y}^{2}} \\
 & \dfrac{180+5p}{16+p}=\dfrac{25p}{16+p} \\
 & 180+5p=25p \\
 & 180=20p \\
 & p=9
\end{align}$
> The value of $p$ is $9$ when $4{{x}^{2}}+p{{y}^{2}}=45$ and ${{x}^{2}}-4{{y}^{2}}=5$ cut orthogonally.

Option ‘E’ is correct

Note: If two curves are orthogonal then it means that they intersect each other at right angles.