Question:
The equation of one of the common tangents to the parabola $y^2 = 8x$ and $x^2 + y^2 - 12x + 4 = 0$ is
The equation of one of the common tangents to the parabola $y^2 = 8x$ and $x^2 + y^2 - 12x + 4 = 0$ is
Updated On: Jul 14, 2022
- $ y = - x + 2 $
- $ y = x - 2 $
- $ y = x + 2 $
- None of these
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The Correct Option is C
Solution and Explanation
Any tangent to parabola $y^2 = 8x$ is $y = mx + \frac{2}{m}$ ......(i)
It touches the circle $x^2 + y^2 - 12x + 4 = 0$ .
if the length of perpendicular from the centre $(6, 0)$ is equal to radius $\sqrt{32}$
$\therefore \, \, \frac{6m+ \frac{2}{m}}{\sqrt{m^{2} + 1 }} = \pm\sqrt{32} $
$ \Rightarrow \left(3m + \frac{1}{m}\right)^{2} = 8\left(m^{2} + 1\right)$
$ \Rightarrow \left(3m^{2 } + 1\right)^{2} = 8 \left(m^{4} + m^{2}\right) $
$\Rightarrow m^{4} - 2m^{2} + 1 = 0\Rightarrow m = \pm1 $
Hence, the required tangents are $y = x + 2$ and $y = -x - 2$.
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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.