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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Mathematics
Chapter
Parabola

BITSAT 2022 · Mathematics · Parabola

If the straight line y=m x+c touches the parabola y^2-4 a x+4 a^3=0...

If the straight line y=mx+cy=m x+c touches the parabola y2−4ax+4a3=0y^2-4 a x+4 a^3=0, then cc is

Options

  1. Aam+ama m+\frac{a}{m}
  2. Bam−ama m-\frac{a}{m}
  3. Cam+a2m\frac{a}{m}+a^2 m
  4. Dam−a2m\frac{a}{m}-a^2 m

Correct answer

D. am−a2m\frac{a}{m}-a^2 m

Step-by-step solution

Solving the given equations, (mx+c)2=4ax−4a3⇒m2x2+2mc\cdotx+c2=4ax−4a3⇒m2x2+(2mc−4a)x+c2+4a3=0\begin{aligned}&(m x+c)^2=4 a x-4 a³ \\ & \Rightarrow m² x²+2 m c \cdotx+c²=4 a x-4 a³ \\ & \Rightarrow m² x²+(2 m c-4 a) x+c²+4 a³=0 \end{aligned}Sincethestraightlinetouchestheparabolaatapoint,so,thediscriminant=0the straight line touches the parabola at a point, so, the discriminant=0⇒(2mc−4a)2−4m2(c2+4a3)=0⇒4m2c2−16amc+16a2−4m2c2−16a3m2=0⇒−mc+a−a2m2=0⇒mc=a−a2m2⇒c=am−a2m\begin{aligned}& \Rightarrow \quad(2 m c-4 a)^2-4 m²\left(c^2+4 a³\right)=0 \\ & \Rightarrow 4 m² c²-16 a m c+16 a²-4 m² c²-16 a³ m²=0 \\ & \Rightarrow-m c+a-a² m²=0 \\ & \Rightarrow m c=a-a² m² \Rightarrow c=\frac{a}{m}-a^2 m \end{aligned}$
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