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Standard XII

Mathematics

Equation of Tangent at a Point (x,y) in Terms of f'(x)

Find the x ...

Question

Find the $x$ intercept of the parabola $y = 3 x^{2} - 2 x - 8$ .

\frac{1}{3}

and

- 25

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\frac{4}{3}

and

- 2

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- \frac{4}{3}

and

2

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- \frac{4}{3}

and

- 2

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Solution

The correct option is B $- \frac{4}{3}$ and $2$
To find the $x$ - intercept, we need to equate $y$ to $0$ .
Doing that, we get $0 = 3 x^{2} - 2 x - 8$
$\Rightarrow x = \frac{2 \pm \sqrt{4 + 96}}{6}$
$\Rightarrow x = \frac{2 \pm 10}{6}$
$\Rightarrow x = \frac{- 4}{3}, 2$

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Similar questions

Q. The area included between the parabolas y² = 4x and x² = 4y is (in square units)
(a) 4/3
(b) 1/3
(c) 16/3
(d) 8/3

Q. Find the angle between the following pair of lines:
(i)

\frac{x - 2}{2} = \frac{y - 1}{5} = \frac{z + 3}{- 3}

and

\frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4}

(ii)

\frac{x}{2} = \frac{y}{2} = \frac{z}{1}

and

\frac{x - 5}{4} = \frac{y - 2}{1} = \frac{z - 3}{8}

Q. Statement- $1$ : An equation of a common tangent to the parabola

y^{2} = 16 \sqrt{3} x

and the ellipse

2 x^{2} + y^{2} = 4

y = 2 x + 2 \sqrt{3}

.

Statement- $2$ : If the line

y = m x + \frac{4 \sqrt{3}}{m}, (m \neq 0)

is a common tangent to the parabola

y^{2} = 16 \sqrt{3} x

and the ellipse

2 x^{2} + y^{2} = 4

, then

m

satisfies

m^{4} + 2 m^{2} = 24

Q. Statement-

1

: An equation of a common tangent to the parabola

y^{2} = 16 \sqrt{3} x

and the ellipse

2 x^{2} + y^{2} = 4

y = 2 x + 2 \sqrt{3}

Statement-

2

: If the line

y = m x + \frac{4 \sqrt{3}}{m}, (m \neq 0)

is a common tangent to the parabola

y ² = 16 \sqrt{3} x

and the ellipse

2 x^{2} + y^{2} = 4

then

m

satisfies:

m^{4} + 2 m^{2} = 24.

Find x, if

$(i) {(\frac{1}{4})}^{- 4} \times {(\frac{1}{4})}^{- 8} = {(\frac{1}{4})}^{- 4 x} (i i) {(\frac{- 1}{2})}^{- 19} \div {(\frac{- 1}{2})}^{8} = {(\frac{- 1}{2})}^{- 2 x + 1} (i i i) {(\frac{3}{2})}^{- 3} \div {(\frac{3}{2})}^{5} = {(\frac{3}{2})}^{2 x + 1} (i v) {(\frac{2}{5})}^{- 3} \times {(\frac{2}{5})}^{15} = {(\frac{2}{5})}^{2 + 3 x} (v) {(\frac{5}{4})}^{- x} \div {(\frac{5}{4})}^{- 4} = {(\frac{5}{4})}^{5} (v i) {(\frac{8}{3})}^{2 x + 1} \times {(\frac{8}{3})}^{5} = {(\frac{8}{3})}^{x + 2}$

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Find the x intercept of the parabola y=3x2−2x−8.

The correct option is B −43 and 2To find the x- intercept, we need to equate y to 0.Doing that, we get 0=3x2−2x−8⇒x=2±√4+966⇒x=2±106⇒x=−43,2

Find the $x$ intercept of the parabola $y = 3 x^{2} - 2 x - 8$ .

The correct option is B $- \frac{4}{3}$ and $2$
To find the $x$ - intercept, we need to equate $y$ to $0$ .
Doing that, we get $0 = 3 x^{2} - 2 x - 8$
$\Rightarrow x = \frac{2 \pm \sqrt{4 + 96}}{6}$
$\Rightarrow x = \frac{2 \pm 10}{6}$
$\Rightarrow x = \frac{- 4}{3}, 2$