Question

Find the value of P for which the curves ${x^2} = 9p\left( {9 - y} \right)$ and ${x^2} = p\left( {y + 1} \right)$ cut each other at right angles.

Answer 1

Hint: The curves cut each other at right angle, therefore the tangent at the point of intersection are perpendicular to each other.
That is, ${m_1} \times {m_2} = - 1$
Where ${m_1}$ is the slope of the tangent to the curve ${x^2} = 9p\left( {9 - y} \right)$
And ${m_2}$ is the slope of the tangent to the curve ${x^2} = p\left( {y + 1} \right)$

Complete step by step answer:
Given, the equations of the curves are
${x^2} = 9p\left( {9 - y} \right)$ → (1) and
${x^2} = p\left( {y + 1} \right)$ → (2)
Let us first find the point of intersection of the two curves.
Solving equation (1) and (2) simultaneously, we get
$
  9p\left( {9 - y} \right) = p\left( {y + 1} \right) \\
   \Rightarrow 81 - 9y = y + 1 \\
   \Rightarrow 10y = 80 \\
   \Rightarrow y = 8 \\
 $
Hence, equation (1)
$
   \Rightarrow {x^2} = 9p\left( {9 - 8} \right) \\
   \Rightarrow {x^2} = 9p \\
   \Rightarrow x = \pm 3\sqrt p \\
 $
Therefore, given curves intersect at points $\left( { \pm 3\sqrt p ,8} \right)$ that is, at points A $\left( {3\sqrt p ,8} \right)$ and B $\left( { - 3\sqrt p ,8} \right)$.
Now, we have to calculate their slope by differentiating their equations.
Considering equation (1), we get
$
  {x^2} = 9p\left( {9 - y} \right) \\
   \Rightarrow \dfrac{{{x^2}}}{{9p}} = 9 - y \\
   \Rightarrow y = 9 - \dfrac{{{x^2}}}{{9p}} \\
 $
On differentiating both sides with respect to x, we get
$\dfrac{{dy}}{{dx}} = \dfrac{{ - 2x}}{{9p}}$ → (3)
∴From equation (3), we get
$
  \dfrac{{{x^2}}}{p} = y + 1 \\
   \Rightarrow y = \dfrac{{{x^2}}}{p} - 1 \\
 $
Again, on differentiating with respect to x, we get
$\dfrac{{dy}}{{dx}} = \dfrac{{2x}}{p}$ → (4)
∴ Slope of the tangent to the curve (1) at point A is
$
  {m_1} = \dfrac{{ - 2 \times 3\sqrt p }}{{9p}} \\
   = \dfrac{{ - 2}}{{3\sqrt p }} \\
 $
And the slope of the tangent to the curve (2) at point A is
$
  {m_2} = \dfrac{{2 \times 3\sqrt p }}{p} \\
   = \dfrac{6}{{\sqrt p }} \\
 $
Since the given curves (1) and (2) cut at right angles,
Therefore,
$
  {m_1} \times {m_2} = - 1 \\
   \Rightarrow \dfrac{{ - 2}}{{3\sqrt p }} \times \dfrac{6}{{\sqrt p }} = - 1 \\
   \Rightarrow \dfrac{4}{p} = 1 \\
   \Rightarrow p = 4 \\
 $

Note:
 Some students might raise questions for not considering the intersection point B $\left( { - 3\sqrt p ,8} \right)$ for evaluating the slopes. However, we get the exact value of P (that is, $p = 4$) while considering point B as well.