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Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is :
Two parabolas with a common vertex and with axes along x-axis and y-axis, respectively, intersect each other in the first quadrant. If the length of the latus rectum of each parabola is 3, then the equation of the common tangent to the two parabolas is :
Updated On: Oct 10, 2024
- $4(x + y) + 3 = 0$
- $3(x + y) + 4 = 0$
- $8(2x + y) + 3 = 0$
- $x + 2y + 3 = 0$
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The Correct Option is A
Solution and Explanation
Equation two parabola are $Y^2 = 3x $ and $x^2 = 3y$
Let equation of tangent to $y^2 = 3x$ is y = mx $+ \frac{3}{4m}$
is also tangent to $x^2 = 3y $
$\Rightarrow x^{2}=3mx+\frac{9}{4m}$
$\Rightarrow 4mx^{2}-12m^{2}x-9=0$ have equal roots
$\Rightarrow D = 0$
$\Rightarrow 144 m^{4}=4 \left(4m\right)\left(-9\right)$
$\Rightarrow m^{4}+m=0\Rightarrow m=-1$
Hence common tangent is $y =-x-\frac{3}{4}$
$4\left(x + y\right) + 3 = 0$
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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.
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