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Consider the parabola ${y^2} = 4ax$ and ${x^2} = 4by$. The straight line ${{\text{b}}^{\frac{1}{3}}}y + {a^{\frac{1}{3}}}x + {a^{\frac{2}{3}}}{b^{\frac{2}{3}}} = 0$

a. Touches ${y^2} = 4ax$

b. Touches ${{x}^2} = 4by$

c. Intersects both parabolas in real point.

d. Touch first and intersect others

Last updated date: 15th Jul 2024
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Hint: Find tangents of both parabolas and equate the constate parts. Substitute the calculated m in any of the tangents, you’ll get the answer.

The tangent the parabola ${y^2} = 4ax$ is

$y = mx + \dfrac{a}{m}......................\left( 1 \right)$

The tangent of the parabola ${x^2} = 4ay$ is

$y = mx - b{m^2}...................\left( 2 \right)$

Let the tangent is common to both the parabolas, therefore the constant part should be equal.

$\Rightarrow \dfrac{a}{m} = - b{m^2}$

$\Rightarrow {m^3} = - \dfrac{a}{b}$

$\Rightarrow m = {\left( { - \dfrac{a}{b}} \right)^{\dfrac{1}{3}}} \Rightarrow - {\left( {\dfrac{a}{b}} \right)^{\dfrac{1}{3}}}$

Now put this value in any equation

$\Rightarrow$ from equation (2)

$y = mx - b{m^2}$

$y = - {\left( {\dfrac{a}{b}} \right)^{\dfrac{1}{3}}}x - b{\left( { - {{\left( {\dfrac{a}{b}} \right)}^{\dfrac{1}{3}}}} \right)^2}$

$y = - {\left( {\dfrac{a}{b}} \right)^{\dfrac{1}{3}}}x - {b^{\dfrac{1}{3}}}{a^{\dfrac{2}{3}}}$

$y{b^{\dfrac{1}{3}}} = - {a^{\dfrac{1}{3}}}x - {b^{\dfrac{2}{3}}}{a^{\dfrac{2}{3}}}$

$\Rightarrow y{b^{\dfrac{1}{3}}} + {a^{\dfrac{1}{3}}}x + {b^{\dfrac{2}{3}}}{a^{\dfrac{2}{3}}} = 0$

Which is your required straight line, hence the given straight line is a common tangent to both of the parabolas. So, the straight line touches both the parabola.

$\Rightarrow$ Option (a) and (b) both are correct.

NOTE: - In this type of questions write the tangent of the parabolas and then simplify using common tangent property.