Consider the equation of curve y-x2-3x-3 and x3-Then equation of normal at point- where its ordinate and abscissa are equal- is -2 marks-

Y=x^2 3x+3 When ordinate is equal to abscissa, we have y=x nbsp; ⇒ x=x^2 3x+3 ⇒ x^2 4x+3=0 ⇒ x=1,3 Since x3, we have x=1 Now, dydx=2x 3 Slope of tangent at x ...

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Standard XII

Mathematics

Angle between Two Lines

Consider the ...

Question

Consider the equation of curve $y = x^{2} - 3 x + 3$ and $x \neq 3.$
Then equation of normal at point, where its ordinate and abscissa are equal, is

[2 marks]

x - y = 2

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x + y = 2

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y = x

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y = - x

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Solution

The correct option is C $y = x$
$y = x^{2} - 3 x + 3$
When ordinate is equal to abscissa,
we have $y = x$
$\Rightarrow x = x^{2} - 3 x + 3$
$\Rightarrow x^{2} - 4 x + 3 = 0$
$\Rightarrow x = 1, 3$
Since $x \neq 3,$ we have $x = 1$

Now, $\frac{d y}{d x} = 2 x - 3$
Slope of tangent at $x = 1,$
$\frac{d y}{d x} {∣ ∣ ∣}_{x = 1} = - 1$
$∴$ Slope of normal $= 1$
Equation of normal at $(1, 1)$ is
$y - 1 = 1 (x - 1)$
$\Rightarrow y = x$

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Consider the equation of curve y=x2−3x+3 and x≠3. Then equation of normal at point, where its ordinate and abscissa are equal, is [2 marks]

The correct option is C y=xy=x2−3x+3 When ordinate is equal to abscissa, we have y=x ⇒x=x2−3x+3 ⇒x2−4x+3=0 ⇒x=1,3 Since x≠3, we have x=1 Now, dydx=2x−3 Slope of tangent at x=1, dydx∣∣∣x=1=−1 ∴ Slope of normal =1 Equation of normal at (1,1) is y−1=1(x−1) ⇒y=x

Consider the equation of curve $y = x^{2} - 3 x + 3$ and $x \neq 3.$
Then equation of normal at point, where its ordinate and abscissa are equal, is

[2 marks]