Let A (0,1), B(1,1) and C(1, 0) be the mid points of the sides of a triangle with incentre at the point D. If the focus of the parabola y2 = 4ax passing through D is (α+β2, 0), where α and βare rational numbers, then αβ2 is equal to

Incentre of given triangle =42+22,44+22⇒y2=4ax44+222=4a44+22⇒a=14+22 Focus (a,0)=14+22,0=4−228,0=(α+β2, 0)α=48=12,β=−14αβ2=12−142=162=8

Let A (0,1), B(1,1) and C(1, 0) be the mid points of the sides of a triangle with incentre at the point D. If the focus of the parabola y2 = 4ax passing through D is (α+β2, 0), where α and βare rational numbers, then αβ2 is equal to

Incentre of given triangle =42+22,44+22⇒y2=4ax44+222=4a44+22⇒a=14+22 Focus (a,0)=14+22,0=4−228,0=(α+β2, 0)α=48=12,β=−14αβ2=12−142=162=8

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Let A (0,1), B(1,1) and C(1, 0) be the mid points of the sides of a triangle with incentre at the point D. If the focus of the parabola y² = 4ax passing through D is $(α + β \sqrt{2}, 0)$ , where $α and β$ are rational numbers, then $\frac{α}{β^{2}}$ is equal to

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Detailed Solution

Incentre of given triangle $= (\frac{4}{2 + 2 \sqrt{2}}, \frac{4}{4 + 2 \sqrt{2}})$
$\Rightarrow y^{2} = 4 ax$
${(\frac{4}{4 + 2 \sqrt{2}})}^{2} = 4 a (\frac{4}{4 + 2 \sqrt{2}}) \Rightarrow a = \frac{1}{4 + 2 \sqrt{2}}$
$Focus (a, 0)$
$\begin{array}{l} = (\frac{1}{4 + 2 \sqrt{2}}, 0) \\ = (\frac{4 - 2 \sqrt{2}}{8}, 0) = (α + β \sqrt{2}, 0) \\ α = \frac{4}{8} = \frac{1}{2}, β = - \frac{1}{4} \\ \frac{α}{β^{2}} = \frac{\frac{1}{2}}{{(\frac{- 1}{4})}^{2}} = \frac{16}{2} = 8 \end{array}$