The value of - c - such that the line joining 0-3-5-2 is a tangent to y - c - x -1 is -A- -4B- 2C- 16D- 4

The correct option is D 4 Equation is x + y = 3 ⇒ (x + 1) (3 x) = c x^2 + 3 + 2x = c x^2+3+2x=c b^2=4ac ⇒ c=4 ...

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Standard XII

Mathematics

Equation of Normal for General Equation of a Circle

The value of ...

Question

The value of ‘c’ such that the line joining (0, 3), (5, –2) is a tangent to

y= $\frac{c}{x + 1}$ is ------

-4

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Solution

The correct option is B
4

Equation is x + y = 3

$\Rightarrow (x + 1) (3 - x) = c - x^{2} + 3 + 2 x = c x^{2} + 3 + 2 x = c b^{2} = 4 a c \Rightarrow c = 4$

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The value of ‘c’ such that the line joining (0, 3), (5, –2) is a tangent to y=cx+1is ------

The correct option is B 4 Equation is x + y = 3 ⇒(x+1) (3−x)=c−x2+3+2x=cx2+3+2x=cb2=4ac⇒ c=4

The value of ‘c’ such that the line joining (0, 3), (5, –2) is a tangent to

y= $\frac{c}{x + 1}$ is ------

The correct option is B
4

Equation is x + y = 3

$\Rightarrow (x + 1) (3 - x) = c - x^{2} + 3 + 2 x = c x^{2} + 3 + 2 x = c b^{2} = 4 a c \Rightarrow c = 4$