Question

If the equation of two curves are ${{y}^{2}}=4x$ and ${{x}^{2}}=4y$ intersects at two points P and Q where Q is other than the origin then,
(A) The angle of intersection of two curves at P is ${{90}^{\circ }}$.
(B) The angle of intersection at point Q is ${{\tan }^{-1}}\left( \dfrac{4}{3} \right)$.
(C) Two curves are orthogonal to each other.
(D) Area of triangle formed by the equation of common tangent of the given two curves with coordinate axis is $\dfrac{1}{2}$.

Answer 1

Check each option one – by – one. For option (A), observe the angle between the tangents of the given curves from the graph, at point P. Solve the two equations to get the coordinates of P and Q.
For option (B), find the slope of tangents of the given curves at point Q with the help of differentiation. Assume the slopes as ‘${{m}_{1}}$’ and ‘${{m}_{2}}$’ and apply the formula: - Angle of intersection = ${{\tan }^{-1}}=\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$.
For option (C), check whether at the point of intersection of the curves if they are having their tangents perpendicular to each other or not. If they are perpendicular then the curves are orthogonal otherwise not.
For option (D), assume the equation of common tangent as $y=mx+c$, where m = slope and c = intercept on y – axis. Substitute the value of y in both the curves and form two quadratic equations in x. Put the value of their discriminant equal to 0 and solve for the values of ‘m’ and ‘c’. Substitute these obtained values in $y=mx+c$. Once the equation of the common tangent is determined, find the coordinates of the points where it is cutting x and y – axis. Use the formula: - Area of triangle = $\dfrac{1}{2}\times$ Base $\times$ Height to get the answer.

Complete step by step answer:
We have been provided with two curves ${{y}^{2}}=4x$ and ${{x}^{2}}=4y$ and we have to check the options.
Now, let us solve the two equations to obtain the coordinates of P and Q.
$\Rightarrow {{y}^{2}}=4x$ - (1)
$\Rightarrow {{x}^{2}}=4y$ - (2)
Squaring equation (2), we get,
$\Rightarrow {{x}^{4}}=16{{y}^{2}}$ - (3)
Substituting the value of ${{y}^{2}}$ from equation (1) in equation (3), we get,

& \Rightarrow {{x}^{4}}=16\times 4x \\\ & \Rightarrow {{x}^{4}}=64x \\\ & \Rightarrow {{x}^{4}}-64x=0 \\\ & \Rightarrow x\left( {{x}^{3}}-{{4}^{3}} \right)=0 \\\ \end{aligned}$$ Substituting each term equal to 0, we get, $$\Rightarrow x=0$$ or $$\left( {{x}^{3}}-{{4}^{3}} \right)=0$$ $$\Rightarrow x=0$$ or $${{x}^{3}}={{4}^{3}}$$ $$\Rightarrow x=0$$ or $$x=4$$ At x = 0, $$\Rightarrow y=0$$ [using equation (2)] Also, at x = 4, $$\Rightarrow y=4$$ [using equation (2)] So, the coordinates of point of intersection are (0, 0) and (4, 4). Now, it is given that Q is other than the origin, that means P is the origin. So, P = (0, 0) and Q = (4, 4). The required graph will look like: -  Now we will check the options one – by – one. (i) For option (A), Here, we have to determine the angle of intersection of the curves at P (0, 0), i.e. angle between the slope of their tangent at P. Clearly, we can see that x – axis is the tangent to the curve, $${{x}^{2}}=4y$$ at P and y –axis is the tangent to the curve $${{y}^{2}}=4x$$ at P. Now, slope of tangent to $${{x}^{2}}=4y$$ is 0 because it is parallel to x – axis while slope of tangent to $${{y}^{2}}=4x$$ is $$\infty $$ because it is parallel to the y – axis. Since, the two axis are perpendicular to each other, therefore their tangents are perpendicular to each other. So, the angle between the tangents is $${{90}^{\circ }}$$. Therefore, angle of intersection at point P is $${{90}^{\circ }}$$. Hence, option (A) is correct. (ii) For option (B), Here, we have to determine the angle of intersection at point Q (4, 4). Let us assume the slope of tangent to the curves $${{y}^{2}}=4x$$ and $${{x}^{2}}=4y$$ at point Q is ‘$${{m}_{1}}$$’ and ‘$${{m}_{2}}$$’ respectively. Now, we know that, $$\dfrac{dy}{dx}$$ = slope. So, let us differentiate the two curves. First let us consider, $${{y}^{2}}=4x$$. On differentiating, we get, $$\begin{aligned} & \Rightarrow 2y\times \dfrac{dy}{dx}=4\times \dfrac{dx}{dx} \\\ & \Rightarrow 2\times y\times {{m}_{1}}=4 \\\ & \Rightarrow {{m}_{1}}=\dfrac{4}{2y} \\\ \end{aligned}$$ The given point is Q (4, 4) so, $$\Rightarrow {{m}_{1}}=\dfrac{4}{2\times 4}=\dfrac{1}{2}$$ Now, let us consider $${{x}^{2}}=4y$$. On differentiating, we get, $$\begin{aligned} & \Rightarrow 2x=4\times \dfrac{dy}{dx} \\\ & \Rightarrow 2x=4{{m}_{2}} \\\ & \Rightarrow {{m}_{2}}=\dfrac{x}{2} \\\ \end{aligned}$$ The given point is Q (4, 4) so, $$\Rightarrow {{m}_{2}}=\dfrac{4}{2}=2$$ Applying the formula for angle of intersection, we get, $$\tan \theta =\left| \dfrac{{{m}_{1}}-{{m}_{2}}}{1+{{m}_{1}}{{m}_{2}}} \right|$$, where $$\theta $$ = angle of intersection. $$\begin{aligned} & \Rightarrow \tan \theta =\left| \dfrac{\dfrac{1}{2}-2}{1+\dfrac{1}{2}\times 2} \right| \\\ & \Rightarrow \tan \theta =\left| \dfrac{-\dfrac{3}{2}}{2} \right| \\\ & \Rightarrow \tan \theta =\left| \dfrac{-3}{4} \right| \\\ \end{aligned}$$ Removing the modulus, we get, $$\begin{aligned} & \Rightarrow \tan \theta =\dfrac{3}{4} \\\ & \Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{3}{4} \right) \\\ \end{aligned}$$ So, option (B) is incorrect. (iii) For option (C), Here, we have to check if the curves are orthogonal. Now, two curves are orthogonal if at their point of intersection their tangents are perpendicular to each other. In the above question the points of intersection are P and Q and we already found that at P the tangents are perpendicular to each other. So, the curves are orthogonal at P. Hence, option (C) is correct. (iv) For option (D), Here, we have to find the area of the triangle formed by the equation of tangent and coordinate axis. To find the area, first we need the equation of common tangent. Let us assume the equation of common tangent as $$y=mx+c$$, where m = slope and c = intercept on y – axis. We have to determine ‘m’ and ‘c’. Now, substituting $$y=mx+c$$, in $${{y}^{2}}=4x$$ we get, $$\begin{aligned} & \Rightarrow {{\left( mx+c \right)}^{2}}=4x \\\ & \Rightarrow {{m}^{2}}{{x}^{2}}+{{c}^{2}}+2mcx=4x \\\ & \Rightarrow {{m}^{2}}{{x}^{2}}+x\left( 2mc-4 \right)+{{c}^{2}}=0 \\\ \end{aligned}$$ The above quadratic equation will have only one solution because the line is a tangent to the curve. So, the discriminant must be 0. $$\begin{aligned} & \Rightarrow {{\left( 2mc-4 \right)}^{2}}=4\times {{m}^{2}}\times {{c}^{2}}\left( \because D={{b}^{2}}-4ac \right) \\\ & \Rightarrow 4{{\left( mc-2 \right)}^{2}}=4{{m}^{2}}{{c}^{2}} \\\ & \Rightarrow {{\left( mc-2 \right)}^{2}}={{m}^{2}}{{c}^{2}} \\\ & \Rightarrow {{m}^{2}}{{c}^{2}}+4-4mc={{m}^{2}}{{c}^{2}} \\\ & \Rightarrow 4mc=4 \\\ \end{aligned}$$ $$\Rightarrow mc=1$$ - (4) Similarly, substituting $$y=mx+c$$, in $${{x}^{2}}=4y$$, we get, $$\begin{aligned} & \Rightarrow {{x}^{2}}=4\left( mx+c \right) \\\ & \Rightarrow {{x}^{2}}-mx-4c=0 \\\ \end{aligned}$$ This quadratic equation will also have only one solution because the line is a tangent to the curve. So, the discriminant must be 0. $$\begin{aligned} & \Rightarrow {{\left( -4m \right)}^{2}}=4\times 1\times \left( -4c \right)\left( \because D={{b}^{2}}-4ac \right) \\\ & \Rightarrow 16{{m}^{2}}=-16c \\\ \end{aligned}$$ $$\Rightarrow {{m}^{2}}=-c$$ - (5) Now, substituting the value of c from equation (5) in equation (4), we get, $$\begin{aligned} & \Rightarrow m\times \left( -{{m}^{2}} \right)=1 \\\ & \Rightarrow {{m}^{3}}=-1 \\\ & \Rightarrow m=-1 \\\ \end{aligned}$$ Substituting, m = -1 in equation (4), $$\begin{aligned} & \Rightarrow -1\times c=1 \\\ & \Rightarrow c=-1 \\\ \end{aligned}$$ So, the equation of common tangent is: - $$\begin{aligned} & \Rightarrow y=-x-1 \\\ & \Rightarrow x+y=-1 \\\ \end{aligned}$$ This line represents the graph given below: -  Clearly, it intersects x – axis at A (-1, 0) and y – axis at B (0, -1). So, OAB is the required triangle formed whose area we have to find. Considering OA as the base and OB as the perpendicular of the triangle OAB, we have, Area = $$\dfrac{1}{2}\times $$ Base $$\times $$ Height $$\Rightarrow $$ Area of $$\Delta OAB=\dfrac{1}{2}\times 1\times 1=\dfrac{1}{2}$$ sq. units **So, the correct answer is “Option A, C and D”.** **Note:** One may note that we have checked all the options given in the question. This is because all the given options have different conditions and statements and it is not necessary that only one is correct. It is important to draw the graphs of the equations so that we can easily visualize the important coordinates which are required to solve the question.