Hyperbola

Given, equation of hyperbola is
‌

x2

a2

−‌

y2

b2

=1
Now, the equation of the tangent to be hyperbola is
‌3√2x−4y=12
‌⇒‌‌y=‌

3√2

4

x−3
So, slope of the tangent, mt=‌

3√2

4

and y-intercept is c=−3
Now, the line y=mx+c tangent to the hyperbola is given by
c2‌=a2m2−b2
⇒‌‌(−3)2‌=a2(‌

3√2

4

)2−b2
⇒‌‌9‌=‌

9

8

a2−b2
⇒‌‌9a2−8b2‌=72
Given, eccentricity, e=‌

5

4

‌ So, ‌b2‌=a2(e2−1)
‌=a2((‌

5

4

)2−1)
‌=a2(‌

25

16

−1)=‌

9

16

a2
⇒‌‌16b2‌=9a2
From Eqs. (i) and (ii), we get
b2‌=9‌ and ‌a2=16
∴a2−b2‌=16−9=7