The two curves y2-4 x and x2-y2-6 x-1-0 at the point 1-2 -A- intersect orthogonallyB- intersect at an angle -6C- touch each otherD- intersect at an angle -3

The correct option is C touch each other y^2=4x ⇒ 2y dy/dx=4 ⇒dy/dx=2/y ∴ M_1=1 x^2+y^2 6x+1=0 ⇒ 2x+2y.dy/dx 6=0 ⇒dy/dx=3 x/y ∴ M_2= ( dy/dx )_1,2=1 ∴ M_1=M_2 ...

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Standard XII

Mathematics

Area between Two Curves

The two curve...

Question

The two curves $y^{2}$ = 4x and $x^{2} + y^{2}$ – 6x + 1 = 0 at the point (1, 2):

intersect orthogonally

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intersect at an angle

\frac{π}{3}

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touch each other

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intersect at an angle

\frac{π}{6}

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Solution

The correct option is C touch each other
$y^{2} = 4 x$
$\Rightarrow 2 y \frac{d y}{d x} = 4$
$\Rightarrow \frac{d y}{d x} = \frac{2}{y}$
$∴ M_{1} = 1$
$x^{2} + y^{2} - 6 x + 1 = 0$
$\Rightarrow 2 x + 2 y . \frac{d y}{d x} - 6 = 0$
$\Rightarrow \frac{d y}{d x} = \frac{3 - x}{y}$
$∴ M_{2} = {(\frac{d y}{d x})}_{1, 2} = 1$
$∴ M_{1} = M_{2} \Rightarrow$ curves touch each other

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The two curves y2 = 4x and x2+y2 – 6x + 1 = 0 at the point (1, 2):

The correct option is C touch each other y2=4x ⇒2ydydx=4 ⇒dydx=2y ∴M1=1 x2+y2−6x+1=0 ⇒2x+2y.dydx−6=0 ⇒dydx=3−xy ∴M2=(dydx)1,2=1 ∴M1=M2⇒ curves touch each other

The two curves $y^{2}$ = 4x and $x^{2} + y^{2}$ – 6x + 1 = 0 at the point (1, 2):