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Standard X

Mathematics

Tangent

If point P1,0...

Question

If point $P (1, 0)$ on the curve $y^{2} + x y + 2 x - 2 = 0$ is at the shortest distance from $y = m x + c .$ Then the value of $m$ is

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Solution

The correct option is C $- 2$
For shortest distance, slope of tangent on curve will be equal to the slope of line.
Now, first curve : $y^{2} + x y + 2 x - 2 = 0$
Differentiating both sides w.r.t. $x,$ we get
$2 y y^{'} + x y^{'} + y + 2 = 0$
At, point $(1, 0)$
$0 + y^{'} + 0 + 2 = 0$
$\Rightarrow y^{'} = - 2$
$\Rightarrow {(\frac{d y}{d x})}_{c_{1}} = {(\frac{d y}{d x})}_{c_{2}}$
$\Rightarrow - 2 = m$
$∴ m = - 2$

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Similar questions

Q. If point

P (0, 1)

on the curve

y = x + e^{sin x}

is at the shortest distance from

y = m x + c .

Then the value of

m^{2}

Q. If point

P (h, k)

on the curve

y = x + ln x

is at the shortest distance from straight line

y = 2 x + 3.

Then the value

(h + k)

Q. Let

y = m x + c

is a tangent at point

P

on the curve

x y - 2 x^{2} = 0,

meets the curve again at

Q .

Then which of the following(s) is(are) not correct

Q. If

y - m x + c = 0,

m, c \in R

is a tangent at point

A

to the curve

y^{2} = 2 x^{2} - 2 x,

meeting the curve again at point

B,

then which of the following conditions holds TRUE?

A straight line is said to be an asymptote to the curve y=f(x) if the

⊥

distance of point P(x, y) on the curve from the line tends to zero when

x \to \infty

y \to \infty

or both x & y

\to \infty

(a) Asymptotes parallel to y-axis:

The line

x = k

is asymptote to the curve y=f(x) If

{lim}_{x \to k^{+}} f (x) = \infty

- \infty

{lim}_{x \to k^{-}} f (x) = + \infty

- \infty

b) Asymptotes parallel to x-axis:

The line

y = k

is asymptotes to the curve

y = f (x)

{lim}_{x \to \infty} y = k

{lim}_{x \to - \infty} y = k

Let y=mx+c be an asymptote of the curve y=f(x), then from any point P(x, y) on the curve, the

⊥

distance is

\frac{| y - m x - c |}{\sqrt{1 + m^{2}}}

which

\to 0

as x tends to

+ \infty

+ \infty

Thus

y - m x - c \to 0

x \to + \infty

- \infty

Again

y - m x - c = x (\frac{y}{x} - \frac{c}{x} - m) = 0

x \to \infty

and

y - m x - c \to 0

lim x \to \pm \infty (\frac{y}{x} - \frac{c}{x} - m) = 0

\Rightarrow

lim x \to \pm \infty \frac{y}{x} = m

and

{lim}_{x \to \pm \infty} y - m x - c = 0

\Rightarrow

{lim}_{x \to \pm \infty} y - m x = c

So we come to the fact that if y=mx+c is an asymptote to the curve y=f(x). Then as

x \to + \infty

- \infty

{lim}_{x \to \infty} \frac{y}{x} = m = {lim}_{x \to \infty} \frac{f (x)}{x}

{lim}_{x \to \infty} = y - m x = {lim}_{x \to \infty} {f (x) - m x} = c

On the basis of above information answer the following questions. The vertical & horizontal asymptotes to the curve

y = \frac{e^{x}}{x}

are respectively

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Standard X Mathematics

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If point P(1,0) on the curve y2+xy+2x−2=0 is at the shortest distance from y=mx+c. Then the value of m is

The correct option is C −2For shortest distance, slope of tangent on curve will be equal to the slope of line. Now, first curve : y2+xy+2x−2=0 Differentiating both sides w.r.t. x, we get 2yy′+xy′+y+2=0 At, point (1,0) 0+y′+0+2=0 ⇒y′=−2 ⇒(dydx)c1=(dydx)c2 ⇒−2=m ∴m=−2

If point $P (1, 0)$ on the curve $y^{2} + x y + 2 x - 2 = 0$ is at the shortest distance from $y = m x + c .$ Then the value of $m$ is