The distance of the point from the common tangent , of the curves and is:

5 Explanation: Step 1: Parabola ki equations ko standard form mein likhein Pehli curve: . Yeh jaisa hai, jahan . Doosri curve: . Yeh shifted parabola hai. Step 2: Dono ke liye tangent ki equation likhein Pehle parabola ( ) ke liye tangent: Doosre parabola ( ) ke liye tangent: Shifted parabola ke liye hum ko se replace karenge: Yahan . Step 3: Common tangent ke liye ko compare karein Dono equations ek hi line ko represent karti hain, isliye inka constant part barabaar hona chahiye: Sari terms ko se multiply karne par: (Kyunki sawal mein diya gaya hai) . Step 4: Common tangent ki equation nikalna ki value Equation 1 mein rakhein: L.C.M lene par ( , toh base ban sakta hai): Step 5: Point se perpendicular distance ( ) Final Answer: Distance 5 unit hai. Sahi option (4) hai.

How do I solve JEE 2023 Mathematics questions?

Practice JEE 2023 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

What is the difficulty level of JEE Mathematics questions?

JEE Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Are JEE previous year papers available for free?

Yes. Mathem Solvex provides free access to JEE previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Tip:A–D to answerE for explanationV for videoS to reveal answer

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The distance of the point $(6, -2\sqrt{2})$ from the common tangent $y = mx + c, m > 0$ , of the curves $x = 2y^2$ and $x = 1 + y^2$ is:

Choose the correct answer:

Explanation

Step 1: Parabola ki equations ko standard form mein likhein

Pehli curve: $y^2 = \frac{1}{2}x$ . Yeh $y^2 = 4ax$ jaisa hai, jahan $4a = 1/2 \implies a = 1/8$ .

Doosri curve: $y^2 = x - 1$ . Yeh shifted parabola hai.

Step 2: Dono ke liye tangent ki equation likhein

Pehle parabola ( $y^2 = \frac{x}{2}$ ) ke liye tangent:

$y = mx + \frac{a}{m} \implies y = mx + \frac{1}{8m} \quad \text{---(Equation 1)}$

Doosre parabola ( $y^2 = x - 1$ ) ke liye tangent:

Shifted parabola ke liye hum $x$ ko $(x-1)$ se replace karenge:

$y = m(x - 1) + \frac{a'}{m}$

Yahan $4a' = 1 \implies a' = 1/4$ .

$y = m(x - 1) + \frac{1}{4m} \implies y = mx - m + \frac{1}{4m} \quad \text{---(Equation 2)}$

Step 3: Common tangent ke liye $c$ ko compare karein

Dono equations ek hi line ko represent karti hain, isliye inka constant part barabaar hona chahiye:

$\frac{1}{8m} = -m + \frac{1}{4m}$

Sari terms ko $8m$ se multiply karne par:

$1 = -8m^2 + 2$

$8m^2 = 1 \implies m^2 = 1/8 \implies m = \frac{1}{\sqrt{8}} = \frac{1}{2\sqrt{2}}$

(Kyunki sawal mein $m > 0$ diya gaya hai).

Step 4: Common tangent ki equation nikalna

$m$ ki value Equation 1 mein rakhein:

$y = \frac{1}{2\sqrt{2}}x + \frac{1}{8(1/2\sqrt{2})} = \frac{x}{2\sqrt{2}} + \frac{2\sqrt{2}}{8} = \frac{x}{2\sqrt{2}} + \frac{\sqrt{2}}{4}$

L.C.M lene par ( $2\sqrt{2} \times \sqrt{2} = 4$ , toh base $4\sqrt{2}$ ban sakta hai):

$4\sqrt{2}y = 2x + 2$

$x - 2\sqrt{2}y + 1 = 0$

Step 5: Point $(6, -2\sqrt{2})$ se perpendicular distance ( $d$ )

$d = \frac{|x_1 - 2\sqrt{2}y_1 + 1|}{\sqrt{1^2 + (-2\sqrt{2})^2}}$

$d = \frac{|6 - 2\sqrt{2}(-2\sqrt{2}) + 1|}{\sqrt{1 + 8}}$

$d = \frac{|6 + 8 + 1|}{3} = \frac{15}{3} = 5$

Final Answer:

Distance 5 unit hai. Sahi option (4) hai.

📄 Download JEE PYQ Papers →

📝 Preparation Tips & Articles →