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(i) we have, $y=x^4-6x^3+13x^2-10x+5$

slope of tangent, $\frac{dy}{dx}=4x^3-18x^2+26x-10$

${\frac{dy}{dx}|}_{(x,y)=(0,5)}=-10.$

slope of normal $=\frac{-1}{-10}=\frac{1}{10}$

Hence eqⁿ of tangent at (0, 5) is

$y-5=-10(x-0)\to 10x+y-5=0$

And eqⁿ of normal at (0, 5) is

$y-5=\frac{1}{10}(x-0)$

$\Rightarrow 10y-50=x$

$\Rightarrow x-10y+50=0$

(ii) We have, $y=x^4-6x^3+13x^2-10x+5$

Slope of tangent, $\frac{dy}{dx}=4x^3-18x^2+26x-10.$

${\frac{dy}{dx}|}_{(x,y)=(1,3)}=4{(1)}^3-18{(1)}^2+26(1)-10$

$=4-18+26-10$

= 30 28

= 2

Slope of normal $=-\frac{1}{2}$

Hence eqⁿ of tangent at (1, 3) is

$y-3=2(x-1)$

$\Rightarrow y-3=2x-2$

$\to 2x-y+1=0$

And eqⁿ of normal at (1,3) is

$(y-3)=-\frac{1}{2}(x-1)$

$\Rightarrow 2y-6=-x+1$

$\to x+2y-7=0$

(iii) We have, $y=x^3$

Slope of tangent, $\frac{dy}{dx}=3x^2$

${\frac{dy}{dx}|}_{(1,1)}=3{(1)}^2=3.$

And slope of normal $=-\frac{1}{3}$

Hence, eqn of tangent at (1, 1) is

$y-1=3(x-1)$

$\Rightarrow y-1=3x-3$

$\to 3x-y-2=0$

And eqⁿ of normal at (1,1) is

$y-1=-\frac{1}{3}(x-1)$

$\to 3y-3=-x+1$

$\Rightarrow x+3y-4=0.$

(iv) We have, $y=x^2$

Slope of tangent $\frac{dy}{dx}=2x$

${\frac{dy}{dx}|}_{(0,0)}=0.$

So, eqⁿ of the tangent at (0,0) is

$(y-0)=0(x-0)$

$\Rightarrow y=0.$

ie, x- axis

Hence, the eqⁿ of normal is x = 0 ie, y-axis

(v) We have, $x=costy=sint$

$\frac{dx}{dt}=-sint·\frac{dy}{dt}=cost$

So, slope of tangent $\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{cost}{-sint}=-cott$

${\frac{dy}{dx}|}_{t=\frac{\pi }{4}}=-cot\frac{\pi }{4}=-1.$

And slope of normal $=\frac{-1}{-1}=1$

This is a Matching Type Questions as classified in NCERT Exemplar

(i) → (e); (ii) → (d); (iii) → (c); (iv) → (b); (v) → (f).

Diffrentiating $\frac{x^2}{9}+\frac{y^2}{16}=1.$ wrt. X we get,

$\frac{2x}{9}+\frac{2y}{16}\frac{dy}{dx}=0$

$\Rightarrow \frac{2ydy}{16dx}=-\frac{2x}{9}$

$\Rightarrow \frac{dy}{dx}=\frac{-16x}{9y}$

(i) When the tangent is to x-axis, the slope of tangent is 0

ie, $\frac{dy}{dx}=0$

$\Rightarrow \frac{-16x}{9y}=0$

$\Rightarrow x=0$ putting this in the eqⁿ of curve. We get,

$\to \frac{0^2}{9}+\frac{y^2}{16}=1$

$\Rightarrow y^2=16$

$\Rightarrow y=\pm 4.$

The point at which the tangents are parallel to x-axis are $(0,4)and(0,-4)$

(ii) When the tangent is parallel to y-axis, the slope of the normal is 0.

ie, $\frac{-1}{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dx$}\right.}=0$

$\Rightarrow -\frac{dx}{dy}=0$

$\Rightarrow \frac{9y}{16x}=0$

$\to y=0$ , putting this in the eqⁿ of curve we get,

$\frac{x^2}{9}+\frac{y^2}{16}=1.$

$\to x^2=9$

$\to x=\pm 3$

The point at which the tangents are parallel to y-axis are $(3,0)and(-3,0)$

The given eqⁿ of the curve is $y=\frac{1}{x^2-2x+3}$

Slope of tangent to the curve is $\frac{dy}{dx}=\frac{-1}{\left(x^2-2x+3\right)}=\frac{d}{dx}\left(x^2-2x+3\right)$

$=\frac{-(2x_1-2)}{{\left(x^2-2x+3\right)}^2}$

Given, $\frac{dy}{dx}=0$

$\to \frac{-(2x-2)}{{\left(x^2-2x+3\right)}^2}=0$

$\Rightarrow -2(x-1)=0$

$\Rightarrow x=1$

When $x=1,y=\frac{1}{1^2-2\times 1+3}=\frac{1}{1-2+3}=\frac{1}{2}$

The point of contact of the tangent to the curve is $\left(1,\frac{1}{2}\right)$

The eqⁿ of the line is $y-\frac{1}{2}=0(x-1)$

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

This is a Matching Type Questions as classified in NCERT Exemplar

(i) → (d); (ii) → (e); (iii) → (c); (iv) → (a)

B.N.M Institute of Technology provides Merit Scholarship to students. Under this scholarships, toppers of each branch in every semester get a scholarship of INR 20,000. Students meeting the specified requirements are offered scholarship. 

The given eqⁿ of curve is $y=\frac{1}{x-3}$

Slope of tangent to the curve is $\frac{dy}{dx}=\frac{-1}{{(x-3)}^2}.$

Given,  $\frac{dy}{dx}=2$

$\Rightarrow \frac{-1}{{(x-3)}^2}=2$

$\Rightarrow {(x-3)}^2=-\frac{1}{2}$ which is not possible

we conclude that there is no possible tangent to the given curve with slope = 2.

Students shortlisted through merit lists can visit MIT School of Vedic Sciences, MIT-ADT University for in-person document verification. After successful verification, partial tuition fee payment is required to confirm admission. Only upon completing both these steps is the admission considered officially confirmed for the academic year.

The given eqⁿ of curve is $y=\frac{1}{x-1}$

Slope of tangent to the given curve is $\frac{dy}{dx}=\frac{-1}{{(x-1)}^2}$

Given that, slope of tangent = 1.

$\to \frac{-1}{{(x-1)}^2}=-1.$

$\Rightarrow {(x-1)}^2=1.$

$\Rightarrow x-1=\pm 1$

$\to x=1\pm 1.$

ie, X=1+1$$ or $x=1-1$

$\Rightarrow x=2$ or $x=0$

When $x=2,y=\frac{1}{2-1}=1$

and when $x=0,y=\frac{1}{0-1}=-1.$

Hence, the point of contact of the tangents are $(2,1)and(0,-1)$

The reqd. eqⁿ of line are $y-1=(-1)(x-2)\{\begin{array}{cc}\hfill ?y-y_0=m\left(x-x_0\right)\hfill & \hfill eqn{}^{} of line \hfill \end{array}$

and $y-(-1)=(-1)(x-0).$

$\Rightarrow y-1=-x+2$ and $y+1=-x$

$\to x+y-3=0$ and $x+y+1=0.$

MIT School of Vedic Sciences selects candidates for its BSc programme based primarily on Class 12 board exam performance. Once applications are reviewed, the university publishes up to three merit lists based on academic records. Shortlisted candidates from these lists are notified and invited to proceed with the next step in the admission process.

The given eqⁿ of the curve is $y=x^3-11x+5$

slope of tangent to the curve $\frac{dy}{dx}=3x^2-11$

Then eqⁿ of tangent is $y=x-11$ $\Rightarrow x-y-11=0$ which gives us slope $=\frac{-1}{-1}=1$

So, $3x^2-11=1$

$\Rightarrow 3x^2=1+11=12$

$\Rightarrow x^2=4$

$\Rightarrow x=\pm 2$

When x = 2, $y=2^3-11(2)+5=8-22+5=-9.$

And when x = 2, $y={(-2)}^3-11(-2)+5=-8+22+5=19.$

The point $(2,-9)$ when put into $y=x-11.$ we get

$-9=2-11$

$\Rightarrow -9=-9$ which is true.

and the point $\left(-2,19\right)$ when put into $y=x-11$ gives,

$19=-2-11$

$\Rightarrow 19=-13$ which is not true.

Hence, the required point is $(2,-9)$ 

Candidates who meet course eligibility can proceed with registration and form filling. The students can visit the website. MIT School of Vedic Sciences BSc application should be submitted on or before the deadline. Candidates should fill details as per official documents.

Let the point joining the chord be $(2,0)(4,4)$

Then slope of the chord $=\frac{4-0}{4-2}\left\{?Slope=\frac{y_2-y_1}{x_2-x_1}\right\}$

$=\frac{4}{2}$

= 2

The given eqⁿ of the curve $y={(x-2)}^2$ 

slope of the tangent to the curve $\frac{dy}{dx}=2(x-2).$

Given that, the tangent is parallel to the chord PQ.

slope of tangent = slope of PQ.

$\Rightarrow 2(x-2)=2.$

$\Rightarrow x=1+2$

$\Rightarrow x=3.$

and $y={(3-2)}^2=1^2=1.$

The required point on curve is $(3,1)$

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The given eqⁿ of the curve is $y=x^3-3x^2-9x+7.$

slope of tangent to the given curve, $\frac{dy}{dx}=3x^2-6x-9$

when the tangent is parallel to x-axis $\frac{dy}{dx}=0$

$\Rightarrow 3x^2-6x-9=0$

$\Rightarrow x^2-2x-3=0$

$\Rightarrow x^2+x-3x-3=0$

$\to x(x+1)-3(x+1)=0$

$\to (x+1)(x-3)=0$

$$ x = 3 or x = -1

When x = 3, $y=3^3-3{(3)}^2-9(3)+7=27-27-27+7=-20$

And when x = -1 $y={(-1)}^3-3{(-1)}^2-9(-1)+7=-1-3+9+7=12$

Hence, the required points are $(3,-20)(-1,12)$

The given eqⁿ of the curves are

$x=1-asin\theta y=bco{s}^2\theta$

so,  $\frac{dx}{d\theta }=-acos\theta \frac{dy}{d\theta }=-2bcos\theta sin\theta$

$\therefore \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{-2bcos\theta sin\theta }{acos\theta }=\frac{2b}{a}sin\theta$

Slope of tangent to curve at $\theta =\frac{\pi }{2}$ is ${\frac{dy}{dx}|}_{\theta =\frac{\pi }{2}}$

$=\frac{2b}{a}sin\frac{\pi }{2}$

$=\frac{2b}{a}$

Hence, slope of normal to curve $=\frac{-1}{2b/a}=-\frac{a}{2b}$

The Equation of the given curve are

$x=aco{s}^3\theta y=asi{n}^3\theta$

So,  $\frac{dx}{d\theta }=3aco{s}^2\theta (-sin\theta )=-3aco{s}^2\theta sin\theta .$

and $\frac{dy}{d\theta }=3asi{n}^2\theta cos\theta$

$\therefore \frac{dy}{dx}=\frac{dy/d\theta }{dx/d\theta }=\frac{3asi{n}^2\theta cos\theta }{-3aco{s}^2\theta sin\theta }=-tan\theta$