Question:
If the parabolas $ y^2 = 4x $ and $ x^2 = 32y $ intersect at $ (16, 8) $ at an angle $ \theta $ , then $ \theta $ equals to
If the parabolas $ y^2 = 4x $ and $ x^2 = 32y $ intersect at $ (16, 8) $ at an angle $ \theta $ , then $ \theta $ equals to
Updated On: Jun 14, 2022
- $ tan^{-1} 5/3 $
- $ tan^{-1} 4/5 $
- $ tan^{-1} 3/5 $
- $ \pi/2 $
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The Correct Option is C
Solution and Explanation
Given curves are
$y^2 = 4x$
$\Rightarrow 2y \frac{dy}{dx} = 4 $
$\Rightarrow \left(\frac{dy}{dx}\right)_{\left(16, 8\right)} = \frac{4}{16} $ $\Rightarrow\left(\frac{dy}{dx}\right)_{\left(16, 8\right)} \frac{1}{4} = m_{1} $ (say)
and $x^{2} = 32y $
$ \Rightarrow 2x = 32 \frac{dy}{dx} $
$ \Rightarrow \left(\frac{dy}{dx}\right)_{\left(16, 8\right)} = \frac{2\times16}{32} $ $\Rightarrow\left(\frac{dy}{dx}\right)_{\left(16,8\right)} = 1 = m_{2}$(say)
$ \therefore$ Angle between them, $\theta = tan^{-1}\left(\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right) $
$ = tan^{-1}\left(\frac{1-\frac{1}{4}}{1+1\times\frac{1}{4}}\right) $
$= tan^{-1}\left(\frac{\frac{3}{4}}{\frac{5}{4}}\right)$
$ = tan^{-1}\left(\frac{3}{5}\right)$
$y^2 = 4x$
$\Rightarrow 2y \frac{dy}{dx} = 4 $
$\Rightarrow \left(\frac{dy}{dx}\right)_{\left(16, 8\right)} = \frac{4}{16} $ $\Rightarrow\left(\frac{dy}{dx}\right)_{\left(16, 8\right)} \frac{1}{4} = m_{1} $ (say)
and $x^{2} = 32y $
$ \Rightarrow 2x = 32 \frac{dy}{dx} $
$ \Rightarrow \left(\frac{dy}{dx}\right)_{\left(16, 8\right)} = \frac{2\times16}{32} $ $\Rightarrow\left(\frac{dy}{dx}\right)_{\left(16,8\right)} = 1 = m_{2}$(say)
$ \therefore$ Angle between them, $\theta = tan^{-1}\left(\frac{m_{2}-m_{1}}{1+m_{1}m_{2}}\right) $
$ = tan^{-1}\left(\frac{1-\frac{1}{4}}{1+1\times\frac{1}{4}}\right) $
$= tan^{-1}\left(\frac{\frac{3}{4}}{\frac{5}{4}}\right)$
$ = tan^{-1}\left(\frac{3}{5}\right)$
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Concepts Used:
Parabola
Parabola is defined as the locus of points equidistant from a fixed point (called focus) and a fixed-line (called directrix).
Standard Equation of a Parabola
For horizontal parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A,
- Two equidistant points S(a,0) as focus, and Z(- a,0) as a directrix point,
- P(x,y) as the moving point.
- Let us now draw SZ perpendicular from S to the directrix. Then, SZ will be the axis of the parabola.
- The centre point of SZ i.e. A will now lie on the locus of P, i.e. AS = AZ.
- The x-axis will be along the line AS, and the y-axis will be along the perpendicular to AS at A, as in the figure.
- By definition PM = PS
=> MP2 = PS2
- So, (a + x)2 = (x - a)2 + y2.
- Hence, we can get the equation of horizontal parabola as y2 = 4ax.
For vertical parabola
- Let us consider
- Origin (0,0) as the parabola's vertex A
- Two equidistant points, S(0,b) as focus and Z(0, -b) as a directrix point
- P(x,y) as any moving point
- Let us now draw a perpendicular SZ from S to the directrix.
- Then SZ will be the axis of the parabola. Now, the midpoint of SZ i.e. A, will lie on P’s locus i.e. AS=AZ.
- The y-axis will be along the line AS, and the x-axis will be perpendicular to AS at A, as shown in the figure.
- By definition PM = PS
=> MP2 = PS2
So, (b + y)2 = (y - b)2 + x2
- As a result, the vertical parabola equation is x2= 4by.