Filters

Q&A - Ask Doubts and Get Answers
Q

Solve it, - Co-ordinate geometry - JEE Main-8

The shortest distance between the line $y=x$ and the curve $y^{2}=x-2$ is :

• Option 1)

$2$

• Option 2)

$\frac{7}{8}$

• Option 3)

$\frac{11}{4\sqrt{2}}$

• Option 4)

$\frac{7}{4\sqrt{2}}$

Views

$\\Line\; \; y=x\\\; \; \; \\Curve=y^{2}=x-2$

Find shortest distance

$\\y^{2}=x-2\\\; \; \; \\2y.y{'}=1$

$y'=\frac{1}{2y}$

the tangent at point P is parallel to $x=y$

So $y'=1=\frac{1}{2y}$

$=y=\frac{1}{2}$

putiing the value of $y$ in curve

$=\left ( \frac{1}{2} \right )^{2}=x-2\Rightarrow \frac{1}{4}=x-2$

$x=\frac{1}{4}+2=\frac{9}{4}$

$\\P=\left ( \frac{9}{4},\frac{1}{2} \right )$

Shortest distance between two parallel lines or perpendicular distance from $P\; to\; x=y$

$=\left | \frac{\frac{9}{4}-\frac{1}{2}}{\sqrt{1^{2}+1^{2}}} \right |=\frac{\frac{7}{4}}{\sqrt{2}}=\frac{7}{4\sqrt{2}}$

Option 1)

$2$

Option 2)

$\frac{7}{8}$

Option 3)

$\frac{11}{4\sqrt{2}}$

Option 4)

$\frac{7}{4\sqrt{2}}$

Exams
Articles
Questions