If the line 2 x-3 y=k touches the parabola ^ 2 =6 x, then find the value of k

Correct answer: B. -27 / 4 aligned Given x=(3 y+k)/(2)& . . and y^ 2 =6 x & y^ 2 =6((3 y+k)/(2)) y^ 2 =3(3 y+k) y^ 2 -9 y-3 k=0 & . . aligned If line (1) touches parabola (2) then roots of quadratic equation ( 3 ) is equal (-9)^ 2 =4 × 1 ×(-3 k) k=-27 / 4

BITSAT 2013

Exam: BITSAT
Year: 2013
Subject: Mathematics
Chapter: Parabola

BITSAT 2013 · Mathematics · Parabola

If the line 2 x-3 y=k touches the parabola ^ 2 =6 x, then find the...

If the line

2 \mathrm{x}-3 \mathrm{y}=\mathrm{k}

touches the parabola

\mathrmy^{2}=6 \mathrm{x},

then find the value of

\mathrm{k}

Options

A $-15 / 4$
B $-27 / 4$
C $-1 / 4$
D $-3 / 4$

Correct answer

B. $-27 / 4$

Step-by-step solution

\begin{aligned}\text{ Given }x=\frac{3 y+k}{2}& \ldots \ldots . . \\ \text{\text{ and } }y^{2}=6 x & \\ \Rightarrow y^{2}=6\left(\frac{3 y+k}{2}\right)\\ \Rightarrow y^{2}=3(3 y+k) \Rightarrow y^{2}-9 y-3 k=0 & \ldots \ldots . . \end{aligned}

If line (1) touches parabola (2) then roots of quadratic equation ( 3 ) is equal$\therefore(-9)^{2}=4 \times 1 \times(-3 k) \Rightarrow k=-27 / 4

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