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Question: The \(y\) intercept of the common tangent to the parabola \({y^2} = 32x\) and \({x^2} = 108y\) is? ...

The yy intercept of the common tangent to the parabola y2=32x{y^2} = 32x and x2=108y{x^2} = 108y is?
A.18 - 18
B.12 - 12
C.9 - 9
D.6 - 6

Explanation

Solution

Hint: First of all compare the given equation with the equation of the parabola y2=4ax{y^2} = 4ax to find the value of aa. Then write the equation of the tangent to the parabola y2=4ax{y^2} = 4ax. Since the tangent is common to the parabola x2=108y{x^2} = 108y, substitute the equation of a line to form a quadratic equation in xx.At last, put the discriminant of the equation equals to find the slope of the tangent common to parabola y2=32x{y^2} = 32x and x2=108y{x^2} = 108y. Substitute 0 for xx in the formed equation of a tangent to find yy intercept.

Complete step by step answer:

Compare the given equation of parabola with the standard equation of y2=4ax{y^2} = 4ax

32x=4ax 4a=32 a=8  \Rightarrow 32x = 4ax \\\ \Rightarrow 4a = 32 \\\ \Rightarrow a = 8 \\\
The equation of the tangent to the parabola y2=32x{y^2} = 32x is given by,
y=mx+8m (1)y = mx + \dfrac{8}{m}{\text{ }}\left( 1 \right)
Also, the tangent meets the parabola, x2=108y{x^2} = 108y
On substituting the value of y=mx+8m y = mx + \dfrac{8}{m}{\text{ }} in the equation of x2=108y{x^2} = 108y, we get,
x2=108(mx+8m ) mx2=108m2x+864 mx2108m2x864=0  \Rightarrow{x^2} = 108\left( {mx + \dfrac{8}{m}{\text{ }}} \right) \\\ \Rightarrow m{x^2} = 108{m^2}x + 864 \\\ \Rightarrow m{x^2} - 108{m^2}x - 864 = 0 \\\
If the line y=mx+8m y = mx + \dfrac{8}{m}{\text{ }} is tangent to the parabola then roots of the equation mx2108m2x864=0m{x^2} - 108{m^2}x - 864 = 0must be equal.
Also, we can say that the discriminant of the equation must be 0.
Discriminant of an equation ax2+bx+c=0a{x^2} + bx + c = 0 is D=b24acD = {b^2} - 4ac .
On substituting the values of a=m,b=108m2,c=864a = m,b = - 108{m^2},c = - 864 in the formula of discriminant, D=b24acD = {b^2} - 4ac, we get,
D=(108m2)24(m)(864)D = {\left( { - 108{m^2}} \right)^2} - 4\left( m \right)\left( { - 864} \right)
Equate it to 0 to find the value of mm.
(108m2)24(m)(864)=0 27m3+8=0 m3=827 m=23  \Rightarrow {\left( { - 108{m^2}} \right)^2} - 4\left( m \right)\left( { - 864} \right) = 0 \\\ \Rightarrow 27{m^3} + 8 = 0 \\\ \Rightarrow {m^3} = - \dfrac{8}{{27}} \\\ \Rightarrow m = - \dfrac{2}{3} \\\
On substituting the value of mm in equation (1) we get,
y=(23)x+823 y=2x363 3y=2x36 2x+3y+36=0  \Rightarrow y = \left( { - \dfrac{2}{3}} \right)x + \dfrac{8}{{ - \dfrac{2}{3}}} \\\ \Rightarrow y = \dfrac{{ - 2x - 36}}{3} \\\ \Rightarrow 3y = - 2x - 36 \\\ \Rightarrow 2x + 3y + 36 = 0 \\\
Thus, the equation of tangent is, 2x+3y+36=02x + 3y + 36 = 0
To find the yyintercept of substitute 0 for xx.
2(0)+3y+36=0 3y=36 y=12  \Rightarrow 2\left( 0 \right) + 3y + 36 = 0 \\\ \Rightarrow 3y = - 36 \\\ \Rightarrow y = - 12 \\\
Hence, option B is the correct answer.

Note: The equation of the line in slope-intercept form is, y=mx+cy = mx + c. If the line touches the parabola, then the discriminant of the resulting quadratic equation is 0. We have to substitute xx as 0 to find the yy intercept.