Question:
The line \(2y = 3x + 12\) cuts the parabola \(4y = 3x^2\). Where does the line cut the parabola?
रेखा \(2y = 3x + 12\) परवलय \(4y = 3x^2\) को काटती है। रेखा परवलय को कहाँ काटती है?
Options:
- At (-1, 3) only
- At (4, 12) only
- At both (-2, 3) and (4, 12)
- Neither at (-2, 3) nor (4, 12)
- केवल (-1, 3) पर
- केवल (4, 12) पर
- (-2, 3) और (4, 12) दोनों पर
- न तो (-2, 3) और न ही (4, 12) पर
Correct Answer: c)
Explanation:
From the line, \(2y = 3x + 12 \implies 4y = 6x + 24\).
Equating to the parabola \(4y = 3x^2\):
\[3x^2 = 6x + 24 \implies 3x^2 – 6x – 24 = 0 \implies x^2 – 2x – 8 = 0\]
\[(x – 4)(x + 2) = 0 \implies x = 4 \text{ or } x = -2\]
– For \(x = -2\): \(2y = 3(-2) + 12 = 6 \implies y = 3\), point is \((-2, 3)\).
– For \(x = 4\): \(2y = 3(4) + 12 = 24 \implies y = 12\), point is \((4, 12)\).
Hence, the line cuts at both (-2, 3) and (4, 12).
Hence, the correct option is (c).
रेखा से: \(4y = 6x + 24\)।
परवलय \(4y = 3x^2\) के बराबर रखने पर:
\[3x^2 – 6x – 24 = 0 \implies (x – 4)(x + 2) = 0 \implies x = 4, -2\]
– \(x = -2\) पर \(y = 3\), बिंदु \((-2, 3)\)
– \(x = 4\) पर \(y = 12\), बिंदु \((4, 12)\)
अतः रेखा दोनों बिंदुओं (-2, 3) और (4, 12) पर काटती है।
अतः सही विकल्प (c) है।