Question

The two parabolas $y^2=4x$ and $x^2=4y$ intersect at a point P, whose abscissae is not zero, such that

• A. they both touch each other at P
• B. they cut at right angles at P
• C. the tangents to each curve at P make complementary angles with the x-axis
• D. none of these

Answer 1

The correct option is C the tangents to each curve at P make complementary angles with the x-axis

Given: $y^2=4x......................\left(1\right)$

$x^2=4y.....................\left(2\right)$

Point of interaction of $\left(1\right)$ and $\left(2\right)$,

$=>\frac{x^4}{16}=4x$

$=>x^3=(4{)}^3$

$=>x=4,0$

But abscissa is not zero, $=>x=4,=>y=4$

So, $P:(4,4)$

Tangent to $\left(1\right)$ at $P=>4y=\frac{4(x+4)}{2}$

$=>2y=x+4............\left(3\right)$

Tangent to $\left(2\right)$ at $P=>4x=\frac{4(y+4)}{2}$

$=>2x=y+4...........\left(4\right)$

Slope of $\left(3\right)$ and (4)$,m_1=\frac{1}{2},$ $m_2=2$

So, ${\alpha }_1={tan}^{-1}\frac{1}{2}$ and ${\alpha }_2={tan}^{-1}2$

$=>{\alpha }_1+{\alpha }_2={tan}^{-1}\frac{1}{2}+{tan}^{-1}2$

$=\frac{\pi }{2}.$