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प्रश्न
The slope of tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, –1) is ______.
विकल्प
`22/7`
`6/7`
`(-6)/7`
– 6
उत्तर
The slope of tangent to the curve x = t2 + 3t – 8, y = 2t2 – 2t – 5 at the point (2, –1) is `6/7`.
Explanation:
The given curve is x = t2 + 3t – 8 and y = 2t2 – 2t – 5
`"dx"/"dt"` = 2t + 3 and `"dy"/"dt"` = 4t – 2
∴ `"dy"/"dx" = ("dy"/"dt")/("dx"/"dt")`
= `(4"t" - 2)/(2"t" + 3)`
Now (2, – 1) lies on the curve
∴ 2 = t2 + 3t – 8
⇒ t2 + 3t – 10 = 0
⇒ t2 + 5t –2t – 10 = 0
⇒ t(t + 5) – 2(t + 5) = 0
⇒ (t + 5)(t – 2) = 0
∴ t = 2, t = – 5 and – 1 = 2t2 – 2t – 5
⇒ 2t2 – 2t – 4 = 0
⇒ t2 – t – 2 = 0
⇒ t2 – 2t + t – 2 = 0
⇒ t(t – 2) + 1(t – 2) = 0
⇒ (t + 1)(t – 2) = 0
⇒ t = – 1 and t = 2
So t = 2 is common value
∴ Slope `"dy"/("dx"_(x = 2)) = (4 xx 2 - 2)/(2 xx 2 + 3) = 6/7`
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