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WBJEE 2022 Math Solved Paper

Question : 66 of 75

Marks: +1, -0

The line

y=x+5

touches

Solution:

(A, B, C are correct )

Hint: (A)

y^{2}=20x=4(5)x\;\therefore

Tangent

: y=mx+\;\frac{a}{m}

for

m=1, a=5

\therefore y=x+5

is a tangent to

y^{2}=20x

(B)

9x^{2}+16y^{2}=144

c^{2}=a^{2}m^{2}+b^{2}

y=x+5 \Rightarrow m=1; c=5

\Rightarrow \;\frac{x^{2}}{16}+\;\frac{y^{2}}{9}=1

\therefore 5^{2}=16(1)^{2}+9 \Rightarrow

True

\therefore y=x+5

is a tangent to

9x^{2}+16y^{2}=144

(C)

c^{2}=a^{2}m^{2}-b^{2}

\frac{x^{2}}{29}-\frac{y^{2}}{4}=1 \rightarrow a^{2}=29; b^{2}=4

5^{2}=29(1)^{2}-4 \rightarrow

True

\therefore y=x+5

is a tangent

(D)

\therefore y=x+5

is NOT tangent to

x^{2}+y^{2}=25

(A), (B), (C) are correct.

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