Question

The two parabolas $y^2=4x$ and $x^2=4y$ intersect at a point $P$ at which the abscissa is not equal to zero. Then,

• A. both of them touch at $P$
• B. they cut at right angles at $P$
• C. the tangents at $P$ make complementary angles with the x-axis
• D. None of these

Answer 1

Answer: (C)

the tangents at $P$ make complementary angles with the x-axis

Solving $y^2=4x$ ........(1)
and $x^2=4y$ ....(2)
${\left(\frac{x^2}{4}\right)}^2=4x\Rightarrow \frac{x^4}{16}=4x.$
$\Rightarrow x(x^3-64)=0\Rightarrow x=4\Rightarrow y=4$
$\therefore P$ is a point $(4,4)$
The tangents at $P$ to $\left(1\right)$ and $\left(2\right)$ are $4y=2(x+4)$ and $4x=2(y+4)$
i.e., $x-2y+4=0$ and $2x-y-4=0$.

Then, slope $m_1=2$ and $m_2=\frac{1}{2}$
If ${\theta }_1$ and ${\theta }_2$ are the inclination of these with the $x-$axis

Then,

$\tan ({\theta }_1+{\theta }_2)=\frac{m_1+m_2}{1-m_1{m}_2}$

$\Rightarrow \tan ({\theta }_1+{\theta }_2)=\frac{2+\frac{1}{2}}{1-2.\frac{1}{2}}$
$\Rightarrow {\theta }_1+{\theta }_2=\frac{\pi }{2}$