The two parabola $${y^2} = {\text{ }}4x$$ and $${x^2} = {\text{ }}4y$$ intersects at a point which the abscissa is not equal to zero. Then,A) both of them touch at PB) they cut at right angle at PC) the tangents at P makes complementary angles with the x-axisD) None of these

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Vedantu Content Team · Accepted Answer

Hint: First find the points or coordinates of P then put the value in the given $${y^2} = {\text{ }}4x$$ and $${x^2} = {\text{ }}4y$$ equation and find slope following that find α is the angle and then choose the option. If the parabola is rotated so that its vertex is (h, k) and its axis of symmetry is parallel to the x-axis, it has an equation of $(y-k)^(2)=4p(x-h)$, where the focus is (h + p, k) and the directory is x = h - p.Complete step-by-step answer:We have$${y^2} = {\text{ }}4x$$ –––––– (1)$${x^2} = {\text{ }}4y$$ –––––– (2)\n    \n    \n    \n    \n  We will 1st find the point of intersection.So, point of intersection of (1) and (2) is$$\frac{{{x^2}}}{{16}} = 4x$$$$ \Rightarrow {x^3} = {\text{ }}{\left( 4 \right)^3}$$$$ \Rightarrow $$ x = 4, 0But, Here abscissa is not zero $$ \Rightarrow $$x = 4, y = 4So, P (4, 4)Here the tangent at P from equation (2) = $$ \Rightarrow 4y = \frac{{4(x + 4)}}{2}$$$$ \Rightarrow  2y = x + 4 $$ ––––––– (3)Also, the tangent at P from equation (2) =$$ \Rightarrow 4x = \dfrac{{4(y + 4)}}{2}$$$$ \Rightarrow  2x = y + 4 $$ –––––––– (4)Now, we need to find the slope of equation (3) and equation (4)Therefore:Slope of equation (3) $$ \Rightarrow {m_1} = \frac{1}{2} $$  [∵ 2y = x + 4]$$y = \dfrac{x}{2} + \dfrac{4}{2}$$$$y = \dfrac{1}{2}x + 2$$We know that $$\dfrac{1}{2}$$ here before x is the slope]Similarly the slope of $$\left( 4 \right){\text{ }} = {m_2} = {\text{ }}2$$If $m_1$ and $m_2$ are the inclination of these with the x−axisThen, $tan(m_1+m_2)=\dfrac{(m_1+m_2)}{1-m_1 \times m_2}$$\Rightarrow$ $tan(m_1+m_2)=\dfrac{(1/2+2)}{1-1/2 \times 2}$$\Rightarrow$ $tan(m_1+m_2)=\dfrac{(3/2)}{0}$$\Rightarrow$ $$ m_1+m_2= \frac{\pi }{2}.$$Therefore the tangents to each curve at P make complementary angles with the x-axis.So the correct option is Option (C).Note: In this type of questions we need to solve the question and then choose into the assertion given in the option and selections according to the solution. Falling objects move along parabolic paths. If a is a positive number then the parabola opens upward and if a is a negative number then the parabola opens downward.

The two parabola \[{y^2} = {\text{ }}4x\] and \[{x^2} = {\text{ }}4y\] intersects at a point which the abscissa is not equal to zero. Then,A) both of them touch at PB) they cut at right angle at PC) the tangents at P makes complementary angles with the x-axisD) None of these

The two parabola \[{y^2} = {\text{ }}4x\] and \[{x^2} = {\text{ }}4y\] intersects at a point which the abscissa is not equal to zero. Then,
A) both of them touch at P
B) they cut at right angle at P
C) the tangents at P makes complementary angles with the x-axis
D) None of these